A statistical research yields to the sample proportion of a certain feature in a certain community be 0.80. Find the required sample size for a maximum margin of error of 0.09 for a 90% confidence interval. a. 54 b. 45 c. 35 d. 53

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Answer 1

Answer:

Step-by-step explanation:

5+6


Related Questions

Use the following information to sketch a graph of the original function, f(X) write the equations of any asymptotes. - lim x→[infinity]

f(x)=5 - f ′
(x)>0 on (−2,1)∪(1,[infinity]) - f ′
(x)<0 on (−[infinity],−2) - f ′′
(x)>0 on (−[infinity],−4)∪(1,4) - f ′′
(x)<0 on (−4,−2)∪(−2,1)∪(4,[infinity])

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The equations of the vertical asymptotes can be given as x = -4, -2, and 1. The function f(x) does not have any horizontal asymptotes.

The function, f(x) is given as f(x)=5 - f ′(x)>0 on (−2,1)∪(1,[infinity]) f ′(x)<0 on (−[infinity],−2)f ′′(x)>0 on (−[infinity],−4)∪(1,4)f ′′(x)<0 on (−4,−2)∪(−2,1)∪(4,[infinity])

To sketch the graph of the original function, we have to determine the critical points, intervals of increase and decrease, the local maximum and minimum, and asymptotes of the given function.

Using the given information, we can form the following table of f ′(x) and f ′′(x) for the intervals of the domain.

The derivative is zero at x = -2, 1.

To get the intervals of increase and decrease of the function f(x), we need to test the sign of f ′(x) at the intervals

(−[infinity],−2), (-2,1), and (1,[infinity]).

Here are the results:

f′(x) > 0 on (−2,1)∪(1,[infinity])f ′(x) < 0 on (−[infinity],−2)

As f ′(x) is positive on the intervals (−2,1)∪(1,[infinity]) which means that the function is increasing in these intervals.

While f ′(x) is negative on the interval (−[infinity],−2), which means that the function is decreasing in this interval.

To find the local maximum and minimum, we need to determine the sign of f ′′(x).

f ′′(x)>0 on (−[infinity],−4)∪(1,4)

f ′′(x)<0 on (−4,−2)∪(−2,1)∪(4,[infinity])

We find the inflection points of the function f(x) by equating the second derivative to zero.

f ′′(x) = 0 for x = -4, -2, and 1.

The critical points of the function f(x) are -2 and 1.

The inflection points of the function f(x) are -4, -2, and 1.

Hence, the equations of the vertical asymptotes can be given as x = -4, -2, and 1.The function f(x) does not have any horizontal asymptotes.

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Find the level of confidence assigned to an interval estimate of the mean formed using the following intervals. (Round your answers to four decimal places.)
(a) x − 0.93·σx to x + 0.93·σx
(b) x − 1.67·σx to x + 1.67·σx
(c) x − 2.17·σx to x + 2.17·σx
(d) x − 2.68·σx to x + 2.68·σx

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The level of confidence assigned to the interval estimates are:

(a) 82.89%

(b) 95.45%

(c) 98.48%

(d) 99.63%

To find the level of confidence assigned to an interval estimate of the mean, we need to use the z-table to determine the corresponding z-score for each given interval multiplier.

The level of confidence can be calculated by subtracting the area in the tails from 1 and multiplying by 100%.

(a) x - 0.93·σx to x + 0.93·σx

The interval multiplier is 0.93. Using the z-table, we find the area in the tails corresponding to this value: 0.1711. Therefore, the level of confidence is approximately (1 - 0.1711) * 100% = 82.89%.

(b) x - 1.67·σx to x + 1.67·σx

The interval multiplier is 1.67. Using the z-table, we find the area in the tails corresponding to this value: 0.0455. Therefore, the level of confidence is approximately (1 - 0.0455) * 100% = 95.45%.

(c) x - 2.17·σx to x + 2.17·σx

The interval multiplier is 2.17. Using the z-table, we find the area in the tails corresponding to this value: 0.0152. Therefore, the level of confidence is approximately (1 - 0.0152) * 100% = 98.48%.

(d) x - 2.68·σx to x + 2.68·σx

The interval multiplier is 2.68. Using the z-table, we find the area in the tails corresponding to this value: 0.0037. Therefore, the level of confidence is approximately (1 - 0.0037) * 100% = 99.63%.

In summary, the level of confidence assigned to the interval estimates are:

(a) 82.89%

(b) 95.45%

(c) 98.48%

(d) 99.63%

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INSTRUCTIONS: Choose the letter of the correct answer. 1. What is the order that we must consider in picking u? a. Logarithmic Function - Inverse Trigonometry Function - Trigonometric Function - Algebraic Function - Exponential Function b. Logarithmic Function - Inverse Trigonometry Function - Algebraic Function - Trigonometric Function - Exponential Function C. Inverse Trigonometry Function - Logarithmic Function - Algebraic Function - - Trigonometric Function - Exponential Function |- d. Logarithmic Function - Exponential Function - Inverse Trigonometry Function - Algebraic Function - Trigonometric Function 2. What is the formula for using Integration by Parts? a. fudv=uv-fvdu b. Sudv=uv + ſ vdu c. Sudv=vdu - fuv d. fudv=vdu + fuv 3. Evaluate fx cosx dx a. xsinx — cosxtc b. xsinx + cosx - C C. xcosx sinx + c d. xcosx + sinx + c 4. Evaluate In2x dx a. xln2x2x+c b. xln2x + x + c C. xln2x-x+c d. 2xlnx + x + c 5. Evaluate fx² cosx dx a. x² sinx + 2xcosx-2sinx+c b. x sinx-2xcosx + 2sinx + c C. x² sinx-2xcosx - 2sinx + c d. x²sinx + 2xcosx + 2sinx + c

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The correct order for picking u in Integration by Parts is: Logarithmic Function - Inverse Trigonometry Function - Algebraic Function - Trigonometric Function - Exponential Function.

The formula for using Integration by Parts is: ∫f(x)g(x)dx = f(x)∫g(x)dx - ∫f'(x)∫g(x)dx.

The evaluation of ∫f(x)cos(x)dx gives the answer xsin(x) - cos(x) + C.

The evaluation of ∫ln(2x)dx gives the answer xln(2x) - x + C.

The evaluation of ∫f(x)²cos(x)dx gives the answer x²sin(x) - 2xcos(x) - 2sin(x) + C.

When using Integration by Parts, it is important to choose the correct order for picking u. The correct order is determined by the acronym "LIATE," which stands for Logarithmic Function, Inverse Trigonometry Function, Algebraic Function, Trigonometric Function, and Exponential Function. Among the given options, the correct order is (a) Logarithmic Function - Inverse Trigonometry Function - Trigonometric Function - Algebraic Function - Exponential Function.

Integration by Parts is a technique used to integrate the product of two functions. The formula for Integration by Parts is ∫f(x)g(x)dx = f(x)∫g(x)dx - ∫f'(x)∫g(x)dx. This formula allows us to split the integral into two parts and simplify the integration process.

To evaluate ∫f(x)cos(x)dx, we use Integration by Parts. By choosing f(x) = x and g'(x) = cos(x), we find f'(x) = 1 and g(x) = sin(x). Applying the formula, we get xsin(x) - ∫sin(x)dx, which simplifies to xsin(x) - cos(x) + C.

To evaluate ∫ln(2x)dx, we again use Integration by Parts. By choosing f(x) = ln(2x) and g'(x) = 1, we find f'(x) = 1/x and g(x) = x. Applying the formula, we get xln(2x) - ∫(1/x)x dx, which simplifies to xln(2x) - x + C.

To evaluate ∫f(x)²cos(x)dx, we once again apply Integration by Parts. By choosing f(x) = x² and g'(x) = cos(x), we find f'(x) = 2x and g(x) = sin(x). Applying the formula, we get x²sin(x) - ∫2xsin(x)dx. Integrating ∫2xsin(x)dx leads to -2xcos(x) - 2sin(x) + C. Thus, the final result is x²sin(x) - 2xcos(x) - 2sin(x) + C.

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answer the following, Round final answer to 4 decimal places. a.) Which of the following is the correct wording for the randon variable? r×= the percentage of all people in favor of a new building project rv= the number of people who are in favor of a new building project r N= the number of people polled r×= the number of people out of 10 who are in favor of a new building project b.) What is the probability that exactly 4 of them favor the new building project? c.) What is the probabilitv that less than 4 of them favor the new building project? d.) What is the probabilitv that more than 4 of them favor the new building project? e.) What is the probabilitv that exactly 6 of them favor the new building project? f.) What is the probability that at least 6 of them favor the new building project? 8.) What is the probabilitv that at most 6 of them favor the new building project?

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In this problem, we are dealing with a random variable related to people's opinions on a new building project. We are given four options for the correct wording of the random variable and need to determine the correct one. Additionally, we are asked to calculate probabilities associated with the number of people who favor the new building project, ranging from exactly 4 to at most 6.

a) The correct wording for the random variable is "rv = the number of people who are in favor of a new building project." This wording accurately represents the random variable as the count of individuals who support the project.

b) To calculate the probability that exactly 4 people favor the new building project, we need to use the binomial probability formula. Assuming the probability of a person favoring the project is p, we can calculate P(X = 4) = (number of ways to choose 4 out of 10) * (p^4) * ((1-p)^(10-4)). The value of p is not given in the problem, so this calculation requires additional information.

c) To find the probability that less than 4 people favor the new building project, we can calculate P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3). Again, the value of p is needed to perform the calculations.

d) The probability that more than 4 people favor the new building project can be calculated as P(X > 4) = 1 - P(X ≤ 4) = 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)).

e) The probability that exactly 6 people favor the new building project can be calculated as P(X = 6) using the binomial probability formula.

f) To find the probability that at least 6 people favor the new building project, we can calculate P(X ≥ 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10).

g) Finally, to determine the probability that at most 6 people favor the new building project, we can calculate P(X ≤ 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6).

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A Group of 10 people sits at a circular table to discuss, every time the leader of the group who always organise the discussion sits at the same seat at the table, the other 9 seats are variable, there are 3 persons of the group do not like sitting next to each other and reject to do that.
How many arrangements the group members can sit around the table?

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In this scenario, a group of 10 people is sitting at a circular table for a discussion. One person, the leader, always sits in the same seat, while the other 9 seats are variable. However, there are 3 individuals in the group who do not want to sit next to each other. The task is to determine the number of arrangements for the group members around the table.

To solve this problem, we can break it down into two steps. First, we arrange the 3 individuals who do not want to sit next to each other. This can be done using the principle of permutations without repetition. Since there are 3 individuals to arrange, we have 3! (3 factorial) ways to arrange them.

Next, we arrange the remaining 7 individuals (including the leader) and the empty seats. Since the table is circular, we consider it as a circular permutation. The number of circular permutations for 7 individuals is (7-1)! = 6!.

Finally, we multiply the number of arrangements for the 3 individuals by the number of circular permutations for the remaining 7 individuals. So, the total number of arrangements is 3! * 6!.

In general, for a circular table with n seats and m individuals who do not want to sit next to each other, the number of arrangements would be m! * (n-m)!.

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Which one is correct about positive and negative biases and bias towards zero and bias away from zero? Check all that apply. (Two correct answers.) A positive bias when the true coefficient is negative, is the same as a bias towards zero. A positive bias when the true coefficient is positive, is the same as a bias away from zero. It is impossible to have a positive bias when the true coefficient is negative. A positive bias is the same as a bias towards zero and a negative bias is the same as a bias away from zero.

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The correct statements about positive and negative biases and bias towards zero and bias away from zero are:

1. A positive bias when the true coefficient is negative is the same as a bias towards zero.

2. A positive bias when the true coefficient is positive is the same as a bias away from zero.

Bias refers to the systematic deviation of the estimated coefficient from the true value in statistical analysis. It can be positive or negative, indicating the direction of the deviation, and can be towards zero or away from zero, indicating the magnitude of the deviation.

If the true coefficient is negative and there is a positive bias, it means that the estimated coefficient is consistently overestimating the true value. In this case, the positive bias is towards zero because the estimated coefficient is being pulled closer to zero than the true negative value.

Conversely, if the true coefficient is positive and there is a positive bias, it means that the estimated coefficient is consistently underestimating the true value. In this case, the positive bias is away from zero because the estimated coefficient is being pushed further away from zero than the true positive value.

It is possible to have a positive bias when the true coefficient is negative. This occurs when the estimated coefficient consistently overestimates the magnitude of the negative effect, resulting in a positive bias towards zero.

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5x – 18 > 2(4x – 15).

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The solution to the inequality 5x - 18 > 2(4x - 15) is x < 4.

To solve the inequality 5x - 18 > 2(4x - 15), we can simplify the expression and isolate the variable x.

First, distribute the 2 to the terms inside the parentheses:

5x - 18 > 8x - 30

Next, we want to isolate the x terms on one side of the inequality.

Let's move the 8x term to the left side by subtracting 8x from both sides:

5x - 8x - 18 > -30

Simplifying further, we combine like terms:

-3x - 18 > -30

Now, let's isolate the variable x.

We can start by adding 18 to both sides of the inequality:

-3x - 18 + 18 > -30 + 18

Simplifying further:

-3x > -12

To isolate x, we need to divide both sides of the inequality by -3. However, when we divide by a negative number, we need to flip the inequality sign:

(-3x) / (-3) < (-12) / (-3)

Simplifying gives us:

x < 4.

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A newspaper published an article about a study in which researchers subjected laboratory gloves to stress. Among 225 vinyl gloves, 67% leaked viruses. Among 225 latex gloves, 8% leaked viruses. Using the accompanying display of the technology results, and using a 0.01 significance level, test the claim that vinyl gloves have a greater virus leak rate than latex gloves. Let vinyl gloves be population 1.

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To test the claim that vinyl gloves have a greater virus leak rate than latex gloves, we compare the virus leak rates of 225 vinyl gloves (67% leaked) and 225 latex gloves (8% leaked) using a significance level of 0.01.

To test this claim, we can perform a hypothesis test by setting up the null and alternative hypotheses:

Null Hypothesis (H0): The virus leak rate for vinyl gloves is equal to or less than the virus leak rate for latex gloves.

Alternative Hypothesis (Ha): The virus leak rate for vinyl gloves is greater than the virus leak rate for latex gloves.

Using the provided data, we can calculate the test statistic and p-value to make a decision.

We can use the normal approximation to the binomial distribution since the sample sizes are large enough. The test statistic can be calculated using the formula:

z = (p1 - p2) / sqrt(p * (1 - p) * ((1/n1) + (1/n2)))

where p1 and p2 are the sample proportions (virus leak rates) of vinyl gloves and latex gloves, n1 and n2 are the respective sample sizes, and p is the pooled proportion calculated as (x1 + x2) / (n1 + n2).

Once the test statistic is calculated, we can find the p-value associated with the observed statistic using a standard normal distribution table or statistical software.

If the p-value is less than the significance level of 0.01, we reject the null hypothesis and conclude that there is evidence to support the claim that vinyl gloves have a greater virus leak rate than latex gloves. Otherwise, we fail to reject the null hypothesis.

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8 people are enterd in a race. If there are no ties, in how many ways can the first three places come out?

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Answer:

336 ways

Step-by-step explanation:

The number of ways the first three places can come out can be calculated using the concept of permutations. In this case, we want to find the number of permutations of 8 objects taken 3 at a time, which is denoted as P(8, 3).

The formula for permutations is:

P(n, r) = n! / (n - r)!

where n is the total number of objects and r is the number of objects being selected.

Using this formula, we can calculate:

P(8, 3) = 8! / (8 - 3)!

= 8! / 5!

= (8 * 7 * 6 * 5!) / 5!

= 8 * 7 * 6

= 336

Therefore, there are 336 different ways the first three places can come out in the race.

Suppose you are given the following to equations:

1) 8X + 9Y = 9

2) 5X + 9Y = 7

What is the value of Y that solves these two equations simultaneously? Please round your answer to two decimal places.

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The value of Y that solves the given system of equations simultaneously is approximately 0.41.

8X + 9Y = 9

5X + 9Y = 7

We can use the method of substitution or elimination. Let's use the elimination method to solve for Y:

Multiply equation (1) by 5 and equation (2) by 8 to make the coefficients of Y the same:

40X + 45Y = 45

40X + 72Y = 56

Now, subtract equation (1) from equation (2) to eliminate X:

(40X + 72Y) - (40X + 45Y) = 56 - 45

Simplifying, we have:

27Y = 11

Divide both sides by 27 to solve for Y:

Y = 11/27 ≈ 0.4074 (rounded to two decimal places)

Therefore, the value of Y that solves the given system of equations simultaneously is approximately 0.41.

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The value of Y that solves the given system of equations simultaneously is approximately 0.41.

8X + 9Y = 9

5X + 9Y = 7

We can use the method of substitution or elimination.

Let's use the elimination method to solve for Y:

Multiply equation (1) by 5 and equation (2) by 8 to make the coefficients of Y the same:

40X + 45Y = 45

40X + 72Y = 56

Now, subtract equation (1) from equation (2) to eliminate X:

(40X + 72Y) - (40X + 45Y) = 56 - 45

Simplifying, we have:

27Y = 11

Divide both sides by 27 to solve for Y:

Y = 11/27 ≈ 0.4074 (rounded to two decimal places)

Therefore, the value of Y that solves the given system of equations simultaneously is approximately 0.41.

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Given that v is inversely related to w-5, If v-8 when w8, what is w when v=6?

Answers

Answer:

v = 6, w is equal to 9.

Step-by-step explanation:

We are given that v is inversely related to w - 5. This can be represented mathematically as:

v = k/(w - 5)

where k is a constant of proportionality.

We can use this relationship to find the value of k:

v = k/(w - 5)

v(w - 5) = k

Now we can use the value v = 8 when w = 8 to find k:

8(8 - 5) = k

24 = k

So our equation is:

v = 24/(w - 5)

Now we can use this equation to find w when v = 6:

6 = 24/(w - 5)

w - 5 = 24/6

w - 5 = 4

w = 9

Therefore, when v = 6, w is equal to 9.

You are validating a new depression scale in a sample of 50 homeless adults. In the general population, this scale is normally distributed with the population mean estimated at 35 and the population standard deviation estimated at 8 . What is the standard error of the mean based on general population parameters?

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Standard Error of the Mean (SEM) is the standard deviation of the sample statistic estimate of the population parameter. It is calculated using the formula:SEM = s / sqrt (n) where s is the standard deviation of the sample and n is the sample size.

The sample size in this case is n = 50.

The standard deviation of the population is s = 8. Therefore, the standard error of the mean (SEM) based on the general population parameters is:[tex]SEM = 8 / sqrt (50)SEM = 1.13[/tex]The standard error of the mean (SEM) is 1.13 based on the general population parameters.

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Approximate the relative error in surface area when the edges of a 2x2x2 m² cube are mismeasured by 2 cm. O 0.25 O 0.0025 O 0.01 01

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The given data is a cube with an edge length of 2m and a surface area of 24m².

Want to find out the relative error in surface area when the edges of a cube are mismeasured by 2 cm?

The formula for the surface area of a cube: Surface [tex]Area = 6a²[/tex]where a is the edge lengthThe formula for the relative error isRelative [tex]Error = (Error / Exact value) * 100%Let's[/tex]solve the questionSolution: Given cube edge [tex]length (a) = 2 mExact value of Surface Area of cube = 6a² = 6 × 2² = 24 m²[/tex]Mismeasured edge length [tex](a') = 2 m + 2 cm = 2.02 mLength error (Δa) = |a - a'| = |2 - 2.02| = 0.02[/tex]mExact value of Surface[tex]Area of cube = 6a² = 6 × 2² = 24 m²Approximated Surface Area (A') = 6a'² = 6 × (2.02)² = 24.48 m²[/tex][tex]Surface Area Error (ΔA) = |A' - A| = |24.48 - 24| = 0.48 m²Relative Error = (Error / Exact value) * 100%Relative Error = (0.48/24) * 100%Relative Error = 0.02 * 100%Relative Error = 2%The relative error in surface area is 2%.[/tex]

Therefore, the correct option is 0.02.

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How many strings of 5 upper case letters of the English alphabet
start or end with A? Letters could be repeted.

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There are 913,952 strings of 5 uppercase letters of the English alphabet that start or end with A.

To find the number of strings of 5 uppercase letters of the English alphabet that start or end with A, we can consider the two cases separately: starting with A and ending with A.

Case 1: Starting with A

In this case, we have one fixed letter A at the beginning, and the remaining four letters can be any uppercase letter, including A. So, there are 26 options for each of the remaining four positions, giving us a total of 26^4 possible strings.

Case 2: Ending with A

Similarly, we have one fixed letter A at the end, and the remaining four letters can be any uppercase letter, including A. Again, there are 26 options for each of the remaining four positions, giving us another 26^4 possible strings.

Since the two cases are mutually exclusive, to find the total number of strings, we need to sum the number of strings in each case:

Total number of strings = Number of strings starting with A + Number of strings ending with A

                     = 26^4 + 26^4

                     = 2 * 26^4

                     = 2 * 456,976

                     = 913,952

Therefore, there are 913,952 strings of 5 uppercase letters of the English alphabet that start or end with A.

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Each of the following statements is an attempt to show that a given series is convergent or divergent using the Comparison Test (NOT the Limit Comparison Test.) For each statement, enter C (for "correct") if the argument is valid, or enter I (for "incorrect") if any part of the argument is flawed. (Note: if the conclusion is true but the argument that led to it was wrong, you must enter 1.) In(n) с 1. For all n > 2, > 1/1, and the series Σ diverges, so by the Comparison Test, the n n n series Σ diverges. n C 2. For all n > 1, 1 < and the series Σ " converges, so by the Comparison Test, the 7-n³ n² n² series Σ converges. 3. For all n > 1, 1 < n ln(n) n and the series 2 Σ diverges, so by the Comparison Test, the " n 1 series Σ diverges. n ln(n) In(n) C 4. For all n > 1, 1 1 and the series Σ " converges, so by the Comparison Test, the n² n¹.5 n1.5 series Σ converges. In(n) n² n C 5. For all n > 2, and the series 2 Σ , " n³-4 converges, so by the Comparison Test, n the series Σ converges. n³-4 6. For all n > 2, 1 n²-4 converges, so by the Comparison Test, the n² series Σ converges. 1 n²-4 In(n) n n 7-n³ < n² < and the series Σ " n² Ť

Answers

Incorrect statement

1. For all n > 2, In(n) > 1/1, and the series Σ In(n) diverges, so by the Comparison Test, the series Σ n/n diverges.

3. For all n > 1, 1 < n ln(n) < n, and the series Σ n²/n diverges, so by the Comparison Test, the series Σ n ln(n) diverges.

7. For all n > 2, 1 < 7-n³ < n², and the series Σ n² converges.

1. For all n > 2, In(n) > 1/1, and the series Σ In(n) diverges, so by the Comparison Test, the series Σ n/n diverges.

Response: I (Incorrect)

The argument is flawed. Comparing In(n) to 1/1 does not provide a conclusive comparison for the convergence or divergence of the series Σ In(n).

2. For all n > 1, 1 < 7-n³/n² < n²/n², and the series Σ n²/n² converges, so by the Comparison Test, the series Σ 7-n³ converges.

Response: C (Correct)

3. For all n > 1, 1 < n ln(n) < n, and the series Σ n²/n diverges, so by the Comparison Test, the series Σ n ln(n) diverges.

Response: I (Incorrect)

The argument is flawed. Comparing n ln(n) to n is not a valid comparison for the convergence or divergence of the series Σ n ln(n). Additionally, the series Σ n²/n is not a valid reference series for the comparison.

4. For all n > 1, 1 < In(n) < n, and the series Σ n² converges, so by the Comparison Test, the series Σ In(n) converges.

Response: C (Correct)

5. For all n > 2, 1/n < 1/(n³-4), and the series Σ 1/(n³-4) converges, so by the Comparison Test, the series Σ 1/n converges.

Response: C (Correct)

6. For all n > 2, 1/(n²-4) < 1/n², and the series Σ 1/n² diverges, so by the Comparison Test, the series Σ 1/(n²-4) diverges.

Response: C (Correct)

7. For all n > 2, 1 < 7-n³ < n², and the series Σ n² converges.

Response: I (Incorrect)

The argument does not apply the Comparison Test correctly. To determine the convergence or divergence of the series Σ 7-n³, we need to compare it to a known convergent or divergent series. The given comparison to n² does not provide enough information to make a conclusion.

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Given 1−e−x for x∈(0,[infinity]). Show that this is a valid CDF. Derive the appropriate pdf.

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To prove that the given function is a valid CDF, we need to show the following: It should be non-negative everywhere.

It should be continuous from the right everywhere. It should be non-decreasing everywhere. It should have a limiting value of 0 as x approaches -∞ and 1 as x approaches ∞.

Let us check these properties one by one.

1. Non-negativity of CDF [tex]f(x) = 1-e^{-x}[/tex] is non-negative for all x > 0.f(x) > 0 for all x > 0

Therefore, this property is satisfied.

2. Right continuity of CDF[tex]f(x) = 1-e^{-x}[/tex] is continuous for all x > 0.

Let x0 be an arbitrary point in the domain of the function.

Let us take a sequence xn of values such that xn → x0 as n → ∞.

Then, we need to show that f(xn) → f(x0) as n → ∞.

As the function is defined only for positive values of x, xn > 0 for all n > 0.So, as n → ∞, xn → x0+

Therefore,[tex]lim_{n \to \infty} f(x_n) = \lim_{x \to x_0^+} f(x)= \lim_{x \to x_0^+} (1-e^{-x})=1-e^{-x_0} = f(x_0)[/tex]

Therefore, f(x) is right continuous for all x > 0.

3. Non-decreasing of CDF [tex]f(x) = 1-e^{-x}[/tex] is non-decreasing for all x > 0.

To prove that the function is non-decreasing, we need to show that for all x1 < x2, we have f(x1) ≤ f(x2).

Consider the case when x1 < x2, then[tex]e^{-x1} > e^{-x2}[/tex].

Therefore, [tex]f(x1) = 1-e^{-x1} ≤ 1-e^{-x2} = f(x2)[/tex]

Hence, f(x) is non-decreasing for all x > 0.4. Limiting values of CDF

The limiting value of f(x) as x → ∞ is

[tex]lim_{x \to \infty} f(x) = \lim_{x \to \infty} (1-e^{-x})= 1- \lim_{x \to \infty} e^{-x} = 1 - 0 = 1[/tex]

This property is satisfied.

The limiting value of f(x) as x → -∞ is

[tex]lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} (1-e^{-x})= 1 - \lim_{x \to -\infty} e^{-x} = 1 - \infty = -\infty[/tex]

This property is not satisfied. Therefore, this function is not a valid CDF.

To derive the appropriate PDF, we need to differentiate the CDF [tex]f(x) = 1-e^{-x}.f(x) = 1-e^{-x}[/tex]

Now, we can differentiate both sides with respect to x using the chain rule.

We get:

[tex]f'(x) = (1-e^{-x})' = -(-1)e^{-x} = e^{-x}[/tex]

The PDF is therefore: [tex]f(x) = e^{-x}[/tex] for x > 0 and 0 elsewhere.

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Homothetic preferences and homogeneous utility functions: (a) Prove that a continuous preference relation is homothetic if and only if it can be represented by a utility function that is homogeneous of degree one. (b) Relate this result to the lecture slides (p. 34, preferences and utility, part 2, see Moodle) which say that any preference relation represented by a utility function that is homogeneous of any degree is homothetic (i.e., not necessarily of degree one). How is it possible that both statements are true at the same time?

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The slides' result includes utility functions that are homogeneous of any degree, which covers the case of utility functions that are homogeneous of degree one mentioned in statement (a).

(a) To prove that a continuous preference relation is homothetic if and only if it can be represented by a utility function that is homogeneous of degree one, we need to show the two-way implication. If a preference relation is homothetic, it implies that there exists a utility function that is homogeneous of degree one to represent it. Conversely, if a utility function is homogeneous of degree one, it implies that the preference relation is homothetic.

(b) The result mentioned in the lecture slides states that any preference relation represented by a utility function that is homogeneous of any degree is homothetic. This statement is more general because it includes the case of utility functions that are homogeneous of degree other than one. So, the lecture slides' result encompasses the specific case mentioned in statement (a) as well.

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If, in a one-tail hypothesis test where H 0 is only rejected in the upper tal, the Z ST

AT=−1.01, what is the statistical decision if the null hypothesis is tested at the 0.10 level of significance? a. Compute the p-value for this test. (Round to the nearest four decimal places to the right of the decimal point.) A. -value =2 ∗ (1−NORM-S.DIST (−1.01,1))=1.688 B. p-value = NORM.S.DIST (−1.01,1)=0.1662 C. p-value =2 ∗ NORM.S.DIST (−1.01,1)=0.3125 D. p-value =1-NORM.S.DIST (−1.01,1)=0.8438 b. What is the statistical decision? A. Since the p-value is less than α=0.10, reject H 0
B. Since the p-value is greater than α=0.10, reject H 0
C. Since the p-value is greater than α=0.10, do not rejoct H 0
D. Since the p-value is less than α=0.10, do not rejoct H 0 .

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The answer to this question is option A. Since the p-value is less than α=0.10, reject H0. Given that the Z stat= -1.01 and the null hypothesis is tested at the 0.10 level of significance, we are to determine the statistical decision and compute the p-value for this test.

To compute the p-value for this test, we use the formula, p-value = 2 * (1 - NORM.S.DIST (-1.01, 1))

= 0.1688 (rounded to 4 decimal places).

Therefore, the p-value for this test is 0.1688.To determine the statistical decision, we check if the p-value is less than or greater than α (alpha) which is the level of significance. If the p-value is less than α, we reject H0. If it is greater than α, we fail to reject H0. Given that the p-value is less than α = 0.10, we reject H0.

Therefore, the statistical decision is A. Since the p-value is less than α=0.10, reject H0.

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In a study of red/green color blindness, 650 men and 3000 women are randomly selected and tested. Among the men, 55 have red/green color blindness. Among the women, 6 have red/green color blindness. Test the claim that men have a higher rate of red/green color blindness.
(Note: Type ��p_m ?? for the symbol p m , for example p_mnot=p_w for the proportions are not equal, p_m>p_w for the proportion of men with color blindness is larger, p_m (a) State the null hypothesis: ___________
(b) State the alternative hypothesis: ____________
(c) The test statistic is ______________
(e) Construct the 95 % confidence interval for the difference between the color blindness rates of men and women.
________<(p m ?p w )< _________

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The study aims to test the claim that men have a higher rate of red/green color blindness compared to women. A sample of 650 men and 3000 women was selected, and the number of individuals with red/green color blindness was recorded. The null hypothesis states that the proportions of men and women with color blindness are equal, while the alternative hypothesis suggests that the proportion of men with color blindness is larger. The test statistic can be calculated using the proportions of color blindness in each group. Additionally, a 95% confidence interval can be constructed to estimate the difference in color blindness rates between men and women.

(a) The null hypothesis: p_m = p_w (The proportion of men with color blindness is equal to the proportion of women with color blindness.)

(b) The alternative hypothesis: p_m > p_w (The proportion of men with color blindness is larger than the proportion of women with color blindness.)

(c) The test statistic: z = (p_m - p_w) / sqrt(p_hat * (1 - p_hat) * (1/n_m + 1/n_w))

Here, p_m and p_w represent the proportions of men and women with color blindness, n_m and n_w represent the sample sizes of men and women, and p_hat is the pooled proportion of color blindness.

(e) The 95% confidence interval for the difference between the color blindness rates of men and women can be calculated as:

(p_m - p_w) ± z * sqrt((p_m * (1 - p_m) / n_m) + (p_w * (1 - p_w) / n_w))

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Given a normal population whose mean is 410 and whose standard deviation is 20, find each of the following:
A. The probability that a random sample of 3 has a mean between 422.470766 and 431.015550.
Probability =
B. The probability that a random sample of 16 has a mean between 407.750000 and 419.300000.
Probability =
C. The probability that a random sample of 30 has a mean between 406.604120 and 412.702098.
Probability =

Answers

A.  between 422.470766 and 431.015550 is approximately 0.008.

B.  between 407.750000 and 419.300000 is approximately 0.928.

C. between 406.604120 and 412.702098 is approximately 0.661.

In order to calculate these probabilities, we can use the Central Limit Theorem, which states that the sampling distribution of the sample means will approach a normal distribution, regardless of the shape of the original population, as the sample size increases. We can approximate the sampling distribution of the means using a normal distribution with the same mean as the population mean and a standard deviation equal to the population standard deviation divided by the square root of the sample size.

For part A, we calculate the z-scores corresponding to the lower and upper bounds of the sample mean range, which are (422.470766 - 410) / (20 / sqrt(3)) ≈ 3.07 and (431.015550 - 410) / (20 / sqrt(3)) ≈ 4.42, respectively. We then use a standard normal distribution table or a calculator to find the probability that a z-score falls between these values, which is approximately 0.008.

For part B, we follow a similar approach. The z-scores for the lower and upper bounds are (407.75 - 410) / (20 / sqrt(16)) ≈ -0.44 and (419.3 - 410) / (20 / sqrt(16)) ≈ 1.13, respectively. The probability of a z-score falling between these values is approximately 0.928.

For part C, the z-scores for the lower and upper bounds are (406.60412 - 410) / (20 / sqrt(30)) ≈ -1.57 and (412.702098 - 410) / (20 / sqrt(30)) ≈ 0.58, respectively. The probability of a z-score falling between these values is approximately 0.661.

These probabilities indicate the likelihood of obtaining sample means within the specified ranges under the given population parameters and sample sizes.

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Average salary is $47,500. Normally distributed with a standard
deviation of $5,200. Take a sample of n = 45 h. What is the probability of the average starting salary in your sample being in excess of $50,000 (to 4 decimal places)? A. i. For all possible samples the same size as yours, what percent of average starting salaries would be no more than $46,000 (to 4 decimal places)? A j. For all possible samples the same size as yours, 5% of the average starting salaries will be below what amount (to 2 decimal places with no commas)? \$ A k. For all possible samples the same size as yours, 3% of the average starting salaries will be above what amount (to 2 decimal places with no commas)? \$ A

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In summary, to solve these problems, we need to apply the concept of the central limit theorem and use z-scores to find the corresponding probabilities or percentiles in the normal distribution

To calculate the probability of the average starting salary in the sample being in excess of $50,000, we can use the central limit theorem. Since the sample size is large (n = 45) and the population is normally distributed, the sample means will also be normally distributed. We need to calculate the z-score for the value $50,000 using the formula z = (x - μ) / (σ / √n). Substituting the values, we have z = ($50,000 - $47,500) / ($5,200 / √45). Using the z-table or a calculator, we can find the probability corresponding to the z-score, which represents the probability of the average starting salary being in excess of $50,000.

To determine the percentage of average starting salaries that would be no more than $46,000, we can use the same approach as above. Calculate the z-score using the formula z = ($46,000 - $47,500) / ($5,200 / √45), and then find the corresponding probability. Multiplying the probability by 100 gives us the percentage.

To find the value below which 5% of average starting salaries would fall, we need to find the z-score corresponding to the cumulative probability of 0.05. Using the z-table or a calculator, we can find the z-score and then convert it back to the corresponding salary value using the formula z = (x - μ) / (σ / √n).

To find the value above which 3% of average starting salaries would fall, we follow a similar process. Find the z-score corresponding to a cumulative probability of 0.97 (1 - 0.03), and then convert it back to the salary value.

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Answer all the questions Question One a. Show the equations for calculating 1. Bulk Volume of a reservoir in ft 3 and barrels 2 . Pore Volume of a reservoir in ft 3 and barrel 3 . Hydrocarbon Pore Volume in ft 3 and in barrel.

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The equations for Bulk Volume of a reservoir in ft³ is VB = A*h and in barrels is VB = (A*h) / 5.615. The equations for Pore Volume of a reservoir in ft³ is VP = φ*VB and in barrels is VP = (φ*VB)/5.615. The equations for Hydrocarbon Pore Volume in ft³ is VHC = φ*S*VB and in barrels is VHC = (φ*S*VB)/5.615.

The equations for calculating the bulk volume, pore volume, and hydrocarbon pore volume of a reservoir are as follows:

1. Bulk Volume (VB):

In cubic feet (ft³):VB = A * hIn barrels (bbl):VB = (A * h) / 5.615

Where:

VB = Bulk Volume

A = Cross-sectional area of the reservoir in square feet (ft²)

h = Thickness of the reservoir in feet (ft)

2. Pore Volume (VP):

In cubic feet (ft³):VP = φ * VBIn barrels (bbl):VP = (φ * VB) / 5.615

Where:

VP = Pore Volume

φ = Porosity of the reservoir (dimensionless)

VB = Bulk Volume

3. Hydrocarbon Pore Volume (VHC):

In cubic feet (ft³):VHC = φ * S * VBIn barrels (bbl):VHC = (φ * S * VB) / 5.615

Where:

VHC = Hydrocarbon Pore Volume

φ = Porosity of the reservoir (dimensionless)

S = Saturation of hydrocarbons in the reservoir (dimensionless)

VB = Bulk Volume

The conversion factor from cubic feet (ft³) to barrels (bbl) is 5.615.

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1. What porition in the distribution cormspends to a z-sore of - 1.20: A. Belowe the mean by 1.20 points B. Beiow the mean by a difstance equal to 1.20 stanuard deviations C. Abave the incain try 1.20 points D. Abave the mican try a distance equal to 1.20 standard deviations 2. What zscore corresponds to a score that is above the mean by 2 standard dieviations? A. 1 13. −2 C. 2 D. 41 3. If a student's exam score in Chemistry was the same as the mean score for. the entire Chemistry class of 35 students, what would that stutent's z-score be: A. 2=35.00 8. z=−0.50 C. z=41.00 D. z=0.00 4. For a population with M=75 and 5=5, what is the z - score correspondin g to x=65? A 4
​ =−2.00 Ba 4
​ +1.00 C. +1.50 D. +2.00 5. A zrcore indicates how an individual perfoemed an w test relative to the other people who took the same tent. A. True 9. False 6. Suppose the 3000 students taking Introduction to Prycholody at a lage univera ty all take the same fin al exam. What can you conclude about a rtudeat takug Introduction to Dpychosogy at this univernfy whic taves the finai exam and qas a j-score of +0.80 on the final exam? คi. The rudent's icore was balaw the nuen of the 3000 wiudents. 8. The itudent answe red corsectiy ant 30 quevicions. C. The itudents score harequal to the mears of ait 1000 students. D. The student's score wras above the me in of the 3000 studenta.

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1. The portion in the distribution corresponding to a z-score of -1.20 is option B. Below the mean by a distance equal to 1.20 standard deviations. This is because the z-score measures the number of standard deviations that a given data point is from the mean of the data set.

A z-score of -1.20 means that the data point is 1.20 standard deviations below the mean. 2. The z-score corresponding to a score that is above the mean by 2 standard deviations is option C. 2. This is because the z-score measures the number of standard deviations that a given data point is from the mean of the data set. A score that is 2 standard deviations above the mean corresponds to a z-score of 2.3.

If a student's exam score in Chemistry was the same as the mean score for the entire Chemistry class of 35 students, their z-score would be option D. z = 0.00. This is because the z-score measures the number of standard deviations that a given data point is from the mean of the data set. If the student's score is the same as the mean, their z-score would be zero.4. For a population with M = 75 and

s = 5, the z-score corresponding to

x = 65 is option A.

z = -2.00. This is because the z-score measures the number of standard deviations that a given data point is from the mean of the data set.

Therefore, the z-score can be calculated as follows: z = (x - M) / s

= (65 - 75) / 5

= -2.005. True. A z-score indicates how an individual performed on a test relative to the other people who took the same test.6. The student's score was above the mean of the 3000 students. This is because a z-score of +0.80 means that the student's score was 0.80 standard deviations above the mean of the data set. Therefore, the student performed better than the average student in the class. Option D is the correct answer.

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Some criminologists argue there is a relationship between "impulsivity" and criminal offending. The idea is that impulsive people act on immediate gratification and that since crime involves quick pleasure and only the long-term possibility of any cost (getting caught and punished), it should be highly attractive to them. To test this notion, you take a random sample of 65 people who responded to a personality test showing they were impulsive and a second independent random sample of 80 who indicated by the test that they were not impulsive. Each person was asked to report the number of criminal offenses they have committed in the last year. For the group of 65 impulsive people, they have a mean number of criminal acts of 13.5 with a standard deviation of 4.9. For the group of 80 nonimpulsive people, they have mean number of criminal acts of 10.3 with a standard deviation of 4.0. Test the hypothesis that there is no difference year. For the group of 65 impulsive people, they have a mean number of criminal acts of 13.5 with a standard deviation of 4.9. For the group of 80 nonimpulsive people, they have mean number of criminal acts of 10.3 with a standard deviation of 4.0. Test the hypothesis that there is no difference between the two groups in the number of delinquent acts. Use an alpha of 0.01. Assume that the two population standard deviations are equal (σ1=σ2). What is your alternative hypothesis?
a. H1:μ impulsive ​<μnon_impulsive b. H1:μ impulsive >μnon_impulsive a. H1:μ impulsive ≠μnon_impulsive

Answers

The alternative hypothesis for this problem is given as follows:

H1:μ impulsive ≠ μ non impulsive

How to obtain the null and the alternative hypothesis?

The hypothesis tested for this problem is given as follows:

"There is no difference between the two groups in the number of delinquent acts."

At the null hypothesis, we test if we have no evidence to conclude that the claim is true, hence:

H0: μ impulsive = μ non impulsive

At the alternative hypothesis, we test if we have evidence to conclude that the claim is true, hence:

H1:μ impulsive ≠ μ non impulsive

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q13,1.6
points Question 13, Save O Points: 0 of 1 In its first 10 years a mutual fund produced an average annual refum of 20.37%. Assume that money invested in this fund continues to earn 20.37% compounded an

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By applying the compound interest formula, we can determine the future value of an investment in a mutual fund that produces an average annual return of 20.37% compounded annually.

Assuming a mutual fund produced an average annual return of 20.37% over its first 10 years, and the investment continues to earn the same rate compounded annually, we can calculate the future value of the investment using the compound interest formula. The formula for compound interest is A = P(1 + r/n)^(nt), where A is the future value, P is the principal amount, r is the annual interest rate, n is the number of times the interest is compounded per year, and t is the number of years. By plugging in the given values, we can compute the future value of the investment.

To calculate the future value of the investment, we can use the compound interest formula: A = P(1 + r/n)^(nt). In this case, the principal amount is not specified, so let's assume it to be 1 for simplicity.

Given that the average annual return is 20.37% and the investment continues to earn the same rate compounded annually, we can substitute the values into the formula. The annual interest rate, r, is 20.37% or 0.2037 as a decimal. Since the interest is compounded annually, the compounding frequency, n, is 1. The number of years, t, is not specified, so let's consider a general case.

Plugging these values into the compound interest formula, we have:

A = 1(1 + 0.2037/1)^(1t).

To simplify the expression, we can rewrite it as:

A = (1.2037)^t.

This formula represents the future value of the investment after t years, assuming a 20.37% annual return compounded annually.

The specific number of years is not mentioned in the question, so we cannot calculate the exact future value without that information. However, we can see that the future value will increase exponentially as the number of years increases, reflecting the compounding effect.

For example, if we consider the future value after 20 years, we can calculate:

A = (1.2037)^20 ≈ 8.6707.

This means that the investment would grow to approximately 8.6707 times its original value after 20 years, assuming a 20.37% annual return compounded annually.

In conclusion, by applying the compound interest formula, we can determine the future value of an investment in a mutual fund that produces an average annual return of 20.37% compounded annually. The specific future value depends on the number of years the investment is held, and without that information, we cannot provide an exact value. However, we observe that the investment will grow exponentially over time due to the compounding effect.


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For each of the following integrals, give a power or simple exponential function that if integrated on a similar infinite domain will have the same convergence or divergence behavior as the given integral, and use that to predict whether the integral converges or diverges. Note that for this problem we are not formally applying the comparison test; we are simply looking at the behavior of the integrals to build intuition. (To indicate convergence or divergence, enter one of the words converges or diverges in the appropriate answer blanks.) x²+1 x³+5x+3 da: a similar integrand is so we predict the integral x ₁ dæ: a similar integrand is dx so we predict the integral x+4 f₁dx : a similar integrand is so we predict the integral x³+4 x+3 ₁732733 dx : a similar integrand is so we predict the integral x³+2x²+3

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x²+1 / x³+5x+3 -> diverges

x / x+4 -> converges

x³+4 x+3 / x³+2x²+3 -> diverges

A similar integrand to x²+1 / x³+5x+3 is x / x². The integral of x / x² is ln(x), which diverges as x approaches infinity. Therefore, we can predict that the integral of x²+1 / x³+5x+3 will also diverge.

A similar integrand to x / x+4 is x / x². The integral of x / x² is ln(x), which converges as x approaches infinity. Therefore, we can predict that the integral of x / x+4 will also converge.

A similar integrand to x³+4 x+3 / x³+2x²+3 is x³ / x². The integral of x³ / x² is x², which diverges as x approaches infinity. Therefore, we can predict that the integral of x³+4 x+3 / x³+2x²+3 will also diverge.

The comparison test is a method for comparing the convergence or divergence of two integrals. The test states that if the integral of f(x) converges and the integral of g(x) diverges, then the integral of f(x)/g(x) diverges.

In this problem, we are not formally applying the comparison test. We are simply looking at the behavior of the integrands to build intuition about whether they will converge or diverge. The integrands in the first two problems have a higher degree than the integrands in the last two problems. This means that the integrals in the first two problems will diverge, while the integrals in the last two problems will converge.

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In a large company, 40% of all employees take public transit to work. Part A If 350 employees are selected at random, calculate the probability that more than 43% of the selected employees take public transit to work. Probability = Note: (1) DO NOT NEED to add or subtract 0.5 (2) Keep the standard deviation of the sampling distribution to at least 8 decimal places. (2) Express the probability in decimal form and round it to 4 decimal places (e.g. 0.1234 ).

Answers

The probability that more than 43% of the selected employees take public transit to work is P(Z > 1.377) = 0.0846

Here, we have

In a large company, the probability that an employee takes public transport to work is 40%. The company has a total of employees. If 350 employees are chosen at random, we must first establish that the sample size, n, is big enough to justify the usage of the normal distribution to compute probabilities.

Therefore, it can be stated that n > 10 np > 10, and nq > 10. Where: n = 350

np = 350 × 0.4 = 140

q = 1 − p = 1 − 0.4 = 0.6

np = 350 × 0.4 = 140 > 10

nq = 350 × 0.6 = 210 > 10

Therefore, we can use the normal distribution to compute probabilities.μ = np = 350 × 0.4 = 140σ = sqrt(npq) = sqrt(350 × 0.4 × 0.6) ≈ 8.02Using continuity correction, we obtain:

P(X > 0.43 × 350) = P(X > 150.5) = P((X - μ) / σ > (150.5 - 140) / 8.02) = P(Z > 1.377), where X is the number of employees who use public transport. Z is the standard normal random variable.

The probability that more than 43% of the selected employees take public transit to work is P(Z > 1.377) = 0.0846 (rounded to 4 decimal places).

Therefore, the required probability is 0.0846, which can be expressed in decimal form.

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[tex]\frac{(x-3)(x^2+3x+27)}{(x-9)(x+9)}[/tex]

Answers

First I distributed everything
Then I cancelled out terms and added terms
Then I expanded the X’s and the 27 so I could cancel out the 9x

2x - 1 3x² and 8(x) = cos² (2x) A/ f(x) = = Given that a) Find f(g(x)) (3 marks) π b) Find f(g(6) (2 marks) Leave your answer in exact form

Answers

The composition of functions f(g(x)) is 12x² + 4x - 5. When plugging in 6 for x in f(g(x)), the result is 451. Thus, f(g(6)) = 451.



To find f(g(x)), we substitute g(x) into f(x). Given g(x) = 2x - 1 and f(x) = 3x² + 8(x), we have f(g(x)) = 3(2x - 1)² + 8(2x - 1). Simplifying this expression, we get f(g(x)) = 3(4x² - 4x + 1) + 16x - 8. Expanding further, we have f(g(x)) = 12x² - 12x + 3 + 16x - 8. Combining like terms, f(g(x)) = 12x² + 4x - 5.

To find f(g(6)), we substitute x = 6 into the expression we obtained for f(g(x)). f(g(6)) = 12(6)² + 4(6) - 5 = 12(36) + 24 - 5 = 432 + 24 - 5 = 451.

Therefore, f(g(x)) = 12x² + 4x - 5 and f(g(6)) = 451.

In summary, f(g(x)) represents the composition of functions f and g, where g(x) is substituted into f(x). In this case, the resulting function is 12x² + 4x - 5. When evaluating f(g(6)), we substitute 6 into the expression and find that the value is 451.

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Which of the following is an assumption of ANCOVA? There should be a reasonable correlation between the covariate and dependent variable Homogeneity of regression slopes Covariates must be measured prior to interventions (independent variable) All of the above

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One assumption of ANCOVA (Analysis of Covariance) is that there should be a reasonable correlation between the covariate and the dependent variable.

The assumption of a reasonable correlation between the covariate and the dependent variable is crucial in ANCOVA because the covariate is included in the analysis to control for its influence on the outcome variable. If there is no correlation or a weak correlation between the covariate and the dependent variable, including the covariate in the analysis may not be meaningful or necessary.

The assumption of a reasonable correlation between the covariate and the dependent variable is an important assumption in ANCOVA, as it ensures the covariate has an actual relationship with the outcome variable being examined.

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The market supply for a good is described Qs=20P-200 and the market demand is described by Qd=1000-40P. What is the market equilibrium QUANTITY? is selection a random or non-random process? random non-random 2. What policy interests or goals would the national RepublicanParty need to change in order to generate political support frommembers of racial/ethnic minority groups? Explain. the matching principle in accounting requires the matching of: in a certain year, according to a national Census Bureau, the number of people in a household had a mean of 4.19 and a standard deviation of 2.25. This is based on census information for the population Suppose the Census Bureau d had estimated this mean using a random sample of 325 homes. Suppose the sample had a sample mean of 4.3 and standard deviation of 2.4. Complete parts a through d below a. Identify the random variable X. Indicate whether it is quantitative or categorical What is the random variable X7 A. The mean number of people in a household B. The number of households in the sample C. The number of people in a household D. The number of households in the country Is X quantitative or categorica? A The variable X is quantitative, because each observation is a numerical value that represents a magnitude of inanable B. The variabia X is categorical, because each observation belongs to one of a set of datince C. The variable X is categoncal, because despite the observations being numerical values, the observations do not represent a magnitude of the table D The variabile X is quantitative, because sach observation belongs to one of a set of distinct values h. Describe the center and variability of the population distribution. What would you predict as the shape of the population Exp The center of the population distribution is (Round to two decimal places as needed) The variability of the population distribution is (Found to two decimal places as needed) Sinca the dis Describe the center and valability of the data distribution. What would you predict as the shape of the data distribution Explain pulition distribution is (Round to two decimal places as needed.) The variability of the population distribution is (Round to two decimal places as needed.) Since the distribution is c. Describe the center and variability of the data distribution. What would you predict as the shape of the data distribuon? Explain. The center of the data distribution is (Round to two decimal places as needed.) The variability of the data distribution is (Round to two decimal places as needed.) the distribution is d. Describe the center and variability of the sampling distribution of the sample mean for 325 homes. What would you predict as the shape of the sampling distribution? Explain The center of the sampling distribution of the sample mean is (Round to two decimal places as needed.) Since The variability of the sampling distribution of the sample mean is (Round to two decimal places as needed.) fortunately, arsenic occurs naturally in some ground water. A mean arsenic level of = 8.0 parts per billion (ppb) is considered safe for agricultural use. A well in Texas is used to water cotton crops. This well is tested on a regular basis for arsenic. A random sample of 41 tests gave a sample mean of x = 6.9 ppb arsenic, with s = 2.8 ppb. Does this information indicate that the mean level of arsenic in this well is less than 8 ppb? Use = 0.01.(a) What is the level of significance?State the null and alternate hypotheses.H0: = 8 ppb; H1: 8 ppb H0: = 8 ppb; H1: > 8 ppb H0: > 8 ppb; H1: = 8 ppb H0: = 8 ppb; H1: < 8 ppb H0: < 8 ppb; H1: = 8 ppb(b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution.The Student's t, since the sample size is large and is unknown. The Student's t, since the sample size is large and is known. The standard normal, since the sample size is large and is known. The standard normal, since the sample size is large and is unknown.What is the value of the sample test statistic? (Round your answer to three decimal places.)(c)Estimate the P-value.P-value > 0.100 0.050 < P-value < 0.100 0.010 < P-value < 0.050 0.005 < P-value < 0.010 P-value < 0.005Sketch the sampling distribution and show the area corresponding to the P-value.A plot of the Student's t-probability curve has a horizontal axis with values from 4 to 4. The curve enters the window from the left, just above the horizontal axis, goes up and to the right, changes direction over approximately 0 on the horizontal axis, and then goes down and to the right before exiting the window just above the horizontal axis. The area under the curve between 4 and 2.52 as well as the area under the curve between 2.52 and 4 are both shaded.A plot of the Student's t-probability curve has a horizontal axis with values from 4 to 4. The curve enters the window from the left, just above the horizontal axis, goes up and to the right, changes direction over approximately 0 on the horizontal axis, and then goes down and to the right before exiting the window just above the horizontal axis. The area under the curve between 2.52 and 4 is shaded.A plot of the Student's t-probability curve has a horizontal axis with values from 4 to 4. The curve enters the window from the left, just above the horizontal axis, goes up and to the right, changes direction over approximately 0 on the horizontal axis, and then goes down and to the right before exiting the window just above the horizontal axis. The area under the curve between 4 and 2.52 is shaded.A plot of the Student's t-probability curve has a horizontal axis with values from 4 to 4. The curve enters the window from the left, just above the horizontal axis, goes up and to the right, changes direction over approximately 0 on the horizontal axis, and then goes down and to the right before exiting the window just above the horizontal axis. The area under the curve between 2.52 and 4 is shaded.(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level ?At the = 0.01 level, we reject the null hypothesis and conclude the data are statistically significant. At the = 0.01 level, we reject the null hypothesis and conclude the data are not statistically significant. At the = 0.01 level, we fail to reject the null hypothesis and conclude the data are statistically significant. At the = 0.01 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.(e) Interpret your conclusion in the context of the application.There is sufficient evidence at the 0.01 level to conclude that the mean level of arsenic in the well is less than 8 ppb. There is insufficient evidence at the 0.01 level to conclude that the mean level of arsenic in the well is less than 8 ppb. In about 450 words (you may write more than this without penalty), utilizing the concepts of AER, CRM and CLV, explain how the firm H&M can continue to improve customer experiences and better responds to customer demands. ring general journal se 2-10 30 The company received $5,000 cash in advance of providing su Prepare general journal entries for the following transactions of a new company called Pose-for-Pics. Use the following partial chart of accounts: Cash; Supplies; Prepaid Insurance; Equipment; M. Harris, Capital; Services Revenue; and Utilities Expense. Aug. 1 M. Harris, the owner, invested $6,500 cash and $33,500 of photography equipment in the company. 2 The company paid $2,100 cash for an insurance policy covering the next 24 months. The company purchased supplies for $880 cash. 5 20 The company received $3,331 cash from taking photos for customers. 31 The company paid $675 cash for August utilities. Open a ledger account for Cash in balance column format. Post general journal entries that impact cash from Francia 29 to the ledger account for Cash, and enter the balance after each posting. Ben swims 50,000 yards per week in his practices. Given this amount of training, he will swim the 100-yard butterfly in 42.7 seconds and place 9th in a big upcoming meet. Ben's coach calculates that if Ben increases his practice to 60,000 yards per week. In terms of Ben's time in the big meet, what is his marginal productivity of the number of yards he practices? The marginal product of increasing practioe yards from 50,000 to 60,000 is secondis) (enter a numeric response using a real number rounded to one decimal place), and the marginal product of increasing practioe yards from 60,000 to 70,001o his time will decrease to 37.1 seconds and he will place 7th in the meet. If Ben practioes 70,000 yards per week, his time will be 35.5 seconds and he will win the meet. is second(s). Is there diminishing marginal productivity of practice yards? In terms of Ben's place in the big meet, what is his marginal productivity of the number of yards he practices? The marginal product of increasing practive yards from 50,000 to 60,000 isplace(s), and the marginal product of increasing practice yards from 60,000 to 70,000 ispce(s) Is there diminishing marginal productivity of practice yards? Does Ben's marginal productivity of the number cf yards he practices depend on how he measures his productivity, either place or time, in the big meet? If a participant's total deferrals during a calendar year exceed the maximum allowed: (Select all that apply) * (3 Points) The excess amount is taxable to the participant in the year the excess deferral was made The excess amount will be applied to the deferral limit for next year: The plan must distribute the excess deferrals and earnings back to the participant 24. If an active employee attains age 72 in 2022, which of the following could be true * (3 Points) Plan may require participant to no longer make contributions Plan may require the participant to take their RMD by April 1, 2023 Plan may allow participant to waive RMD while still employed During 2019, Pepsi Co had an average daily cost of goods sold of $13million, average accounts receivable of $91 million, average inventory of $104 million. Whatwas Pepsis average age of inventory for the year?A. 8 daysB. 7 daysC. 1 dayD. 91 daysE. None of the above Assume that the specific production function of the representative firm is:F(K,L,mf)=AKLmfwhere F(.) is the firms production function, K is its capital stock, L is its employmentand m is its holdings of real balances.(i)Derive the relevant demand and supply functionsfor the firm, stating any assumptions that you need to make. Define the user cost ofmoney. [13marks](ii)Show the dependence of marginal productivities of labor and capital on the usercost of real balances.[6marks](iii)Suppose that a financial innovation multiplies the firms marginal productivity of realbalances by . What would be its impact on the firms demand functions for laborandcapital?[6marks The probability is (Round to four decimal places as needed.) cans each week. Set the mean fill level at ounces. (Round to two decimal places as needed.) Since this mean fill level is (1). than 12 ounces, the (2) (1) less (2) customers will be short changed. greater company will lose money from overfilling. During the current year, an ABC company incurred the following costs:R&D services performed by XYZ for ABC: $41,189R&D performed by ABC for QRS: $16,603Design, construction, and testing of pre-production prototypes and models: $45,555Testing in search for new products or process alternatives: $30,851The amount ABC should report as R&D expense is? In the context of corporate culture,________is defined as the gap between individuals at different levels of a hierarchy.a. uncertainty avoidance indexb. power distance indexO c. moral free spaceO d. categorical imperative Please Solve below A. Find a parametrization for the line segment beginning at P and ending at P2. P 1(4, 4, -3) and P 2 0, 4, nd P 2 (0.47) O x = 4t, y = 4t, z = -25 t + 7,0 st1 O 25 x = -4t + 4, y = 4, z = t-3,0 sts1 O x = 4t, y = 4,2 = -25t+7,0sts1 O 25 x = -4t + 4, y = 4t, z = -t-3,0sts 1 B. Find parametric equations for the line described below. The line through the point P(-6, 5, 3) parallel to the vector 2i-6j-6k O x = -2t-6, y = -6t + 5, z = 6t + 3 Ox= -2t + 6, y = 6t - 5, z = -6t - 3 O x = 2t + 6, y = -6t - 5, z = -6t - 3 Ox= 2t -6, y = -6t + 5, z = -6t+3 Australia and Italy have 140 and 150 total resources, respectively. Each country produces only two goods, toilet paper and hand sanitizer. Australia can produce each roll of toilet paper for 14 resources and each bottle of hand sanitizer for 19 resources. Italy can produce each roll of toilet paper for 6 resources and each bottle of hand sanitizer for 38 resources. Assume each country would prefer to consume more toilet paper and hand sanitizer than it currently can in an autarky. Also assume that each country can produce fractional units of each good. Between which range of prices of hand sanitizer (in terms of rolls of toilet paper) will there be gains from trade?Select one:a. Between 1.36 and 6.33b. Between 0.16 and 0.74c. Between 0.43 and 2.00d. Between 0.50 and 2.33e. Between 0.93 and 1.07 A poll finds that 52% of the population approves of the job that the President is doing; the poll has a margin of error of 4% (assuming a 95% degree of confidence). a. What is the 95% confidence interval for the true population proportion that approves of the President's performance? b. What was the minimum size of the sample for this poll? a. Find the 95% confidence interval. A bag contains 30 buttons that are colored either blue, red oryellow. There are the same number of each color (10 each). Atotal 4 buttons are drawn from the bag. Compute the followings:1) Find n().2) The probability that at least 3 of them are red?3) The probability that there is at least one of each color? Control chart systems can operate on two basic methods of measurement. State these two methods and briefly distinguish between them, giving three examples of industrial processes where each might be applied. Specify an appropriate sampling procedure in each case.