The first derivative of at x with respect to x is simply the transpose of the matrix a.The first derivative of xt ax with respect to x is 2ax, since taking the derivative of the product of two vectors involves multiplying one of the vectors by the derivative of the other vector, and in this case the derivative of x is the identity matrix (since x is a vector and not a matrix).The second derivative of xt ax with respect to x is simply the matrix 2a, since the second derivative involves taking the derivative of the first derivative.1. Given the vector x and the symmetric matrix A.
2. We need to find the first and second derivatives of the following expressions:
i. A * x
ii. x^T * A * x
The question involves vectors, symmetric matrices, and derivatives. Let's break it down step-by-step.
i. First derivative of A * x with respect to x:
To find the derivative of A * x with respect to x, we treat A as a constant matrix. The first derivative is simply the matrix A itself.
Answer: The first derivative of A * x with respect to x is A.
ii. First derivative of x^T * A * x with respect to x:
To find the first derivative of this expression, we'll use the following formula for the derivative of a quadratic form:
d/dx (x^T * A * x) = (A + A^T) * x
Since A is a symmetric matrix, A = A^T. Therefore, the formula becomes:
d/dx (x^T * A * x) = 2 * A * x
Answer: The first derivative of x^T * A * x with respect to x is 2 * A * x.
iii. Second derivative of x^T * A * x with respect to x:
The second derivative of x^T * A * x with respect to x is the derivative of the first derivative (2 * A * x) with respect to x. Since A is a constant matrix, the second derivative is zero.
Answer: The second derivative of x^T * A * x with respect to x is 0.
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At Michael’s school, 38% of the students have a pet dog and 24% of the students have a pet cat. Michael found that 11% of the students had both a pet dog and a pet cat. What is the probability that a randomly chosen student at Michael’s school will have a pet dog or a pet cat? A. 51% B. 62% C. 83% D. 40%
Answer:
62%
Step-by-step explanation:
Its addition, 38+24=62
30+20+12=62 to make thing simpler.
Thema deposits 500$ ina savings account with a simple interest rate of 1. 3%. How could you use this information to find the interest she would earn in 4 years and determine the percent change in her savings account
The percent change in her savings account is 5.2%, meaning her savings account has increased by 5.2% due to the interest earned over 4 years.
To find the interest Thema would earn in 4 years, we can use the simple interest formula:
where I is the interest earned, P is the principal (initial deposit), r is the interest rate (as a decimal), and t is the time period in years.
In this case, we have P = $500, r = 0.013, and t = 4. Plugging in these values, we get:
I = [tex]($500) (0.013) (4) = $26[/tex]
So, Thema would earn $26 in interest over 4 years.
To determine the percent change in her savings account, we need to compare the amount she will have after 4 years to the initial deposit. After 4 years, her savings account will have:
A = P + I = $500 + $26 = $526
The percent change in her savings account can be calculated as:
percent change = (new amount - old amount) / old amount x 100%
Substituting the values, we get:
percent change = ($526 - $500) / $500 x 100% = 5.2%
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2. How many ways can you choose 5 players from 14?
3. How many ways can you line up 7 books on a shelf?
4. A test of 10 different types of candy are being tested. If a person is given 6 of the candies to taste how many combinations are possible?
solve using permutation
The number of ways of arrangement is 726, 485, 760 ways, 5040 ways, 210 ways resdpectively
What is permutation?Remember that Permutation is a mathematical technique that determines the number of possible arrangements in a set when the order of the arrangements matters.
How many ways can you choose 5 players from 14
= [tex]14^{} Px_{5}[/tex]
14!/(14-5)!
(14*13*12*11*10*9*8*7*6*5*4*3*2*1)/9*8*7*6*5*4*3*2*1
= 726, 485, 760 ways
3. How many ways can you line up 7 books on a shelf?
= 7!
= 5040 ways
4. 10!/(10-6)!6!
10*9*8*7*6*5*4*3*2*1/6*5*4*3*2*1*4*3*2*1
= 5040/24
= 210 ways
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49 packages are randomly selected from packages received by a parcel service. The sample has a mean weight of 28.6 pounds. Assume that What is the 95% confidence interval for the true mean weight, μ, of all packages received by the parcel service?
A. 27.5 to 29.7 pounds
B. 27.7 to 29.5 pounds
C. 27.6 to 29.6 pounds
D. 27.9 to 29.3 pounds
The correct answer is option C: 27.6 to 29.6 pounds.
To calculate the 95% confidence interval for the true mean weight, μ, of all packages received by the parcel service, we can use the following formula:
CI = X ± (tα/2)(s/√n)
where:
X is the sample mean weight
tα/2 is the t-value from the t-distribution with (n - 1) degrees of freedom and α/2 level of significance (α/2 = 0.025 for a 95% confidence interval)
s is the sample standard deviation of weights
n is the sample size
We are given that the sample size is n = 49 and the sample mean weight is X = 28.6 pounds. We do not have the sample standard deviation, so we will assume it is unknown and use a t-distribution to estimate the standard error of the sample mean.
We want to calculate the 95% confidence interval, so α/2 = 0.025 and the degrees of freedom are (n - 1) = 48. From a t-distribution table, the t-value for a 95% confidence interval with 48 degrees of freedom is approximately 2.01.
Now we need to estimate the sample standard deviation, s, using the sample data. We can use the formula:
s = sqrt[ Σ(xi - X)² / (n - 1) ]
where xi is the weight of the ith package in the sample.
Without the actual data values, we cannot compute the exact value of s, but we can estimate it using the sample variance. We know that:
s² = Σ(xi - X)² / (n - 1) = (49 - 1) / 48 * s²
So, an estimate of s is:
s = sqrt[ (SS / (n - 1)) ] = sqrt[ (Σ(xi - X)² / (n - 1)) ]
where SS is the sum of squares of deviations of the sample data from the mean.
Assuming the sample standard deviation is unknown and using the above formula, we can estimate s as:
s ≈ 2.7 pounds
Substituting the known values into the formula for the confidence interval, we have:
CI = X ± (tα/2)(s/√n)
= 28.6 ± (2.01)(2.7/√49)
= 28.6 ± 0.99
Therefore, the 95% confidence interval for the true mean weight, μ, of all packages received by the parcel service is (28.6 - 0.99, 28.6 + 0.99), or approximately (27.6, 29.6) pounds.
So, the correct answer is option C: 27.6 to 29.6 pounds.
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A population has a mean of 5, with a standard deviation of 1. A sample of 50
items from that population has a mean of 4.5, with a standard deviation of
1.1.
Which equation describes the population parameter?
A. X = 5
OB. = 4.5
C. μ = 5
OD. X = 4.5
Answer:C. μ = 5
Step-by-step explanation:Option A (X = 5) and option D (X = 4.5) are incorrect because X represents a single value or observation, not a population parameter. Option B (μ = 4.5) is also incorrect because the question states that the mean of the population is 4.5, but we are looking for the equation that describes the population parameter which is the true mean of the entire population.
The mean of this data set: 18, 16, _, 9, 12, 23, is 15. What number is missing?
There is a data set with data values 18, 16, _, 9, 12, 23. If mean of data set is 15, then the misssing number is equals to the twelve.
Mean means the average of a data set. It is calculated by adding all numbers together and then dividing the resultant sum by the number of numbers, i.e, count of numbers. It is denoted by
[tex] \bar X .[/tex] Mathematcally formula,
[tex]\bar X = \frac{\sum X_n}{n }[/tex]
Where, Xₙ --> data values or numbers
n --> count of values
We have a data set with data values : 18, 16, _, 9, 12, 23. Mean of this data set = 15
We have to determine the missing number. Let the missing number be x. The sum of numbers = 18 + 16 + x + 9 + 12 + 23 = 78 + x
Count of values = 6
Using above mean formula, [tex] 15 = \frac{ 78+ x}{6}[/tex]
=> 90 = 78 + x
=> x = 90 - 78
=> x = 12
So, required value is 12.
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LOOK AT THE IMAGE AND SOLVE IT FOR BRAINLIEST!!! ASAP!!!
Answer:
A. The correct answer is B, the point where the most profit is made.
B. The correct answer is B, the price per pen where no profit is made.
C.
[tex]m = \frac{120 - 0}{3 - 6} = - \frac{120}{3} = - 40[/tex]
When the price of a pen increases by one dollar, the profit decreases by $40.
D. The domain of this graph given the situation is 0 < x < 6 because there is no profit (there is a loss) beyond those points.
assign sum extra with the total extra credit received given list test grades. full credit is 100, so anything over 100 is extra credit.
To assign the sum of extra credit received to the total extra credit, you need to follow these steps:
1. Make a list of all the test grades, and mark the ones that have extra credit with a "+" sign. For example, if the test scores are: 95, 110+, 80, 120+, 100, you would mark the 110+ and the 120+.
2. Add up all the extra credit marks. In this example, that would be 110 + 120 = 230.
3. Count the number of extra credit marks. In this example, there are two.
4. Multiply the number of extra credit marks by the maximum extra credit value. If full credit is 100, then the maximum extra credit value is the amount that a student can exceed the maximum score by. In this case, that would be 20 (since 120 - 100 = 20). So, 2 x 20 = 40.
5. Add the extra credit value to the total score. In this example, the total score is 505 (95 + 110 + 80 + 120 + 100). So, you would add 40 to that to get 545.
Therefore, the answer is: To assign sum extra with the total extra credit received given list test grades, you need to add the extra credit value to the total score by multiplying the number of extra credit marks by the maximum extra credit value, and then adding that value to the total score.
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Sound City sells the ClearTone-400 satellite car radio. For this radio, historical sales records over the last 100 weeks show 5 weeks with no radios sold, 17 weeks with one radio sold, 17 weeks with two radios sold, 49 weeks with three radios sold, 9 weeks with four radios sold, and 3 weeks with five radios sold. Calculate μx, σx2, and σx, of x, the number of ClearTone-400 radios sold at Sound City during a week using the estimated probability distribution. (Round your answers to 2 decimal places.)
µx
σx2,
σx
The mean is 2.48, the variance is 1.5844, and the standard deviation is 1.26.
To calculate the mean, variance, and standard deviation, we need to first construct a probability distribution table:
where f(x) is the frequency of weeks with x radios sold divided by the total number of weeks (100).
Using this table, we can calculate the mean as:
μx = ∑(x * f(x)) = (00.05) + (10.17) + (20.17) + (30.49) + (40.09) + (50.03) = 2.48
To calculate the variance, we use the formula:
σx2 = ∑((x - μx)2 * f(x)) = (0-2.48)2 * 0.05 + (1-2.48)2 * 0.17 + (2-2.48)2 * 0.17 + (3-2.48)2 * 0.49 + (4-2.48)2 * 0.09 + (5-2.48)2 * 0.03 = 1.5844
Finally, we can calculate the standard deviation as:
σx = √σx2 = √1.5844 = 1.26
Therefore, the mean is 2.48, the variance is 1.5844, and the standard deviation is 1.26.
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How are the hours spent on homework per day related to grade level in school?
Grade in School Hours spent on Homework (per day)
4 1
6 1.5
8 2.5
10 3
12 3.5
a. The higher the grade level in school, the more hours spent on homework.
b. The more hours spent on homework, the lower the grade in school.
c. The higher the grade level in school, the less hours spent on homework.
d. No relationship.
The requried relation is, the higher the grade level in school, the more hours spent on homework. Option A is correct.
From the given data, we can see that the hours spent on homework per day increase as the grade level in school increases. This suggests that there is a positive correlation between grade level and homework hours.
In general, higher grade levels in school involve more advanced coursework and greater academic expectations, which can require more time and effort outside of class to complete homework and study. Therefore, it is reasonable to expect that students in higher grade levels will spend more hours on homework per day compared to students in lower grade levels.
Thus, the requried relation is, the higher the grade level in school, the more hours spent on homework. Option A is correct.
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You pick a card at random, put it back, and then pick another card at random.
2345
What is the probability of picking a 4 and then picking a number less than 5?
Simplify your answer and write it as a fraction or whole number.
There are FOUR 4's in a deck of 52
4/52 = 1/13 chance
Second card less than 5 ?
4 aces 4 two's 4 threes and 4 fours
16/ out of 52 = 16/52 chance = 4/13 chance
1/13 * 4/13 = 4/169 chance of picking as questioned
9. What number exceeds its square by the greatest amount? (2) DO NOT SOLVE. Just write the necessary equations for solving & related let statements.
To find the number that exceeds its square by the greatest amount, we can use these terms: "number" and "square."
Let "n" represent the number.
The square of the number is "n^2."
We are looking for the greatest difference between the number and its square, which can be represented as:
Difference (D) = n - n^2
To find the number that maximizes this difference, we can use calculus to find the critical points (where the derivative is zero or undefined). However, since you asked not to solve it, I'll provide the necessary equations for solving:
1. D = n - n^2
2. Find the derivative of D with respect to n: dD/dn
By following these steps, you can solve for the number that exceeds its square by the greatest amount.
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(b) Prove that if the sequence (4) satisfies lim = L = 0, then a) is unbounded. 71
We have proved that if the sequence (a_n) satisfies lim a_n = L = 0, then (a_n) is unbounded.
To prove that if the sequence (a_n) satisfies lim a_n = L = 0, then (a_n) is unbounded, we will use proof by contradiction.
Assume that (a_n) is bounded. Then, there exists a positive number M such that |a_n| ≤ M for all n in the natural numbers.
Since lim a_n = L = 0, we can choose an ε > 0 such that if n is sufficiently large, then |a_n - L| < ε. In other words, there exists a natural number N such that for all n ≥ N, |a_n - L| < ε.
Consider the case when n ≥ N and a_n > 0 (the case when a_n < 0 is similar). Then, we have:
a_n = L + (a_n - L) > L - ε
Since a_n ≤ M, we have:
0 ≤ a_n < M
Combining these inequalities, we get:
0 ≤ L - ε < a_n < M
This implies that a_n is bounded between two positive numbers, which contradicts the assumption that (a_n) is unbounded. Therefore, our initial assumption that (a_n) is bounded must be false, and hence (a_n) is unbounded when lim a_n = L = 0.
Therefore, we have proved that if the sequence (a_n) satisfies lim a_n = L = 0, then (a_n) is unbounded.
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evaluate the triple integral. 8xyz dv, where t is the solid tetrahedron with vertices (0, 0, 0), (1, 0, 0), (1, 1, 0), and (1, 0, 1) t
To evaluate the triple integral of 8xyz dv over the tetrahedron T with vertices (0, 0, 0), (1, 0, 0), (1, 1, 0), and (1, 0, 1), we need to set up the proper bounds for the integral. We can set up the integral as follows:
∫∫∫_T 8xyz dz dy dx
First, find the equations of the planes that form the tetrahedron. The planes are:
1. x = 1 (constant plane)
2. z = 0 (xy-plane)
3. y = 1 - x (line in the xy-plane)
4. z = 1 - x (line in the xz-plane)
Now, set the bounds for the integral:
x: 0 to 1
y: 0 to 1 - x
z: 0 to 1 - x
So, the triple integral becomes:
∫(0 to 1) ∫(0 to 1-x) ∫(0 to 1-x) 8xyz dz dy dx
Evaluate the innermost integral:
∫(0 to 1) ∫(0 to 1-x) [4xyz(1-x)] dy dx
Now evaluate the second integral:
∫(0 to 1) [8x/3 * (1-x)^3] dx
Finally, evaluate the outermost integral:
[2/15 * (1-x)^5] evaluated from 0 to 1
Plugging in x = 1 gives 0, and plugging in x = 0 gives 2/15.
Therefore, the value of the triple integral is 2/15.
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henry and liani have 200 feet of wood to frame a flower bed. henry wants the bed to be a square, while liani wants it to be a rectangle with dimensions of 55 feet and 45 feet. find the area of henry's flower bed. area
The area of Henry's flower bed would be 2500 square feet.
Let's start by finding the perimeter of Henry's flower bed since we know that he wants it to be a square. If we let s be the length of one side of the square, then the perimeter would be:
4s = 200
Simplifying this equation, we get:
s = 50
So Henry's flower bed will have sides of 50 feet each.
To find the area of the flower bed, we can use the formula:
Area = side^2
So in this case, the area would be:
Area = 50^2 = 2500 square feet
Therefore, the area of Henry's flower bed would be 2500 square feet.
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How many different simple random samples of size 4 can be obtained from a population whose size is 48? The number of simple random samples which can be obtained is ____. (Type a whole number.)
The number of simple random samples of size 4 that can be obtained from a population of size 48 is 194,580.
To find the number of different simple random samples of size 4 that can be obtained from a population whose size is 48, we can use the combination formula. The combination formula is written as:
C(n, k) = n! / (k!(n-k)!)
where n is the population size (48), k is the sample size (4), and ! denotes a factorial (e.g., 5! = 5 × 4 × 3 × 2 × 1).
So, in this case:
C(48, 4) = 48! / (4!(48-4)!)
C(48, 4) = 48! / (4!44!)
C(48, 4) = (48 × 47 × 46 × 45) / (4 × 3 × 2 × 1)
C(48, 4) = 194580
Therefore, the number of simple random samples of size 4 that can be obtained from a population of size 48 is 194,580.
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A colony of bacteria grows so that t days after the start of an experiment, the number of bacteria is n • 2 t/2, when n is the number of bacteria at the start of the expierement if there are 10,000bactrria 6 days after the experiments start what is the value of n
The initial number of bacteria in the colony was approximately 13.5.
The value of n, the number of bacteria at the start of the experiment, can be calculated using the formula n =
[tex](b/2)^(2/t),[/tex]
where b is the number of bacteria at any given time and t is the time in days.
Plugging in the given values, we get: n =
[tex](10,000/2)^(2/6)[/tex]
n =
[tex]2,500^(1/3)[/tex]
n ≈ 13.5. This formula is derived from the fact that the growth of bacteria is often modeled by an exponential function, where the rate of growth is proportional to the current population size.
In this case, the number of bacteria is doubling every 2 days (since [tex]2^(1/2) = 2^(2/4) = 2^(4/8) = ...),[/tex] so we can rewrite the original equation as n • [tex]2^(t/2)[/tex]. Using the given information that there are 10,000 bacteria 6 days after the experiment starts, we can plug in these values and solve for n using the derived formula.
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3) Find the simple interest.
Rafael borrowed $12,000 at 6% interest to be paid back in 7 years. How much interest will that cost him?
Please help asap
Step-by-step explanation:
Simple interest = prt÷100
Here,
Principle = 12000
Rate = 6
Time = 7
So to find the simple interest,
You just apply the formula.
[tex]\frac{p \times r \times t}{100} = \frac{12000 \times 6 \times 7}{100} = 5040[/tex]
Interest = 5040
the cone and cylinder below both have a height of 11 feet. the cone has a radius of 3 feet. the cylinder has a volume of 310.86 cubic feet. complete the statements using 3.14 for . any non-integer answers in this problem should be entered as decimals rounded to the nearest hundredth. the volume of the cone is cubic feet. the radius of the cylinder is feet. the ratio of the volume of the cone to the volume of the cylinder is 1:.
The ratio of the volume of the cone to the volume of the cylinder is approximately 0.33 : 1 (rounded to the nearest hundredth).
The volume of the cone can be calculated using the formula:
V = (1/3)πr^2h
where r is the radius of the cone and h is the height of the cone. Substituting the given values, we get:
V = (1/3)π(3)^2(11) = 103.67 cubic feet
Therefore, the volume of the cone is 103.67 cubic feet (rounded to the nearest hundredth).
To find the radius of the cylinder, we can use the formula for the volume of a cylinder:
V = πr^2h
where r is the radius of the cylinder and h is the height of the cylinder. We are given that the volume of the cylinder is 310.86 cubic feet and that the height is 11 feet, so we can solve for r:
310.86 = πr^2(11)
r^2 = 310.86 / (11π)
r ≈ 2.3 feet
Therefore, the radius of the cylinder is approximately 2.3 feet (rounded to the nearest hundredth).
The ratio of the volume of the cone to the volume of the cylinder is the volume of the cone divided by the volume of the cylinder. Using the values we calculated, we get:
V(cone) / V(cylinder) = 103.67 / 310.86 ≈ 0.33 : 1
Therefore, the ratio of the volume of the cone to the volume of the cylinder is approximately 0.33 : 1 (rounded to the nearest hundredth).
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Select the table of values where the quadratic function changes direction at a different value of x than the others. A. x -3 -2 -1 0 1 2 3 y 4 3 4 7 12 19 28 B. x -3 -2 -1 0 1 2 3 y 1 -1 1 7 17 31 49 C. x -3 -2 -1 0 1 2 3 y 2 3 2 -1 -6 -13 -22 D. x -3 -2 -1 0 1 2 3 y 28 19 12 7 4 3 4 Reset
The solution to the given quadratic equation is x= -3 -2 -1 0 1 2 3 y= 2 3 2 -1 -6 -13 -22.
This table of values contains a quadratic function that changes direction at the value of x = 0. This is different from the other tables of values which all have the quadratic function changing direction at the value of x = -2. The y values in this table of values can be described as an upside-down parabola. At x = 0, the y value is -1 and it decreases as x increases to positive values, and it increases as x decreases to negative values.
Therefore correct answer is C. x -3 -2 -1 0 1 2 3 y 2 3 2 -1 -6 -13 -22.
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Practice problem for module 2 A random sample of 50 students GPA reveals that the mean GPA is 2.8 years with a standard deviation of 0.45 years. (a) Construct a 95% Confidence Interval for the mean lifetime of all LED TV. (b) If we want to be 90% confident, and we want to control the maximum error of estimation to be 0.2, how many more students should be added into the given sample?
(c) Would you conclude that the mean GPA more than 2.5 at 5% level of significance?
a) a 95% confidence interval for the mean lifetime of all LED TV is: (2.664, 2.936)
b)Rounding up, we need to add 28 more students to the sample.
c) The critical value for a one-tailed t-test with 95% confidence and n-1 degrees of freedom is t = 1.729.
Substituting the values, we get:
(a) To construct a 95% confidence interval for the mean lifetime of all LED TV, we can use the formula:
CI = X ± z*(s/√n)
where X is the sample mean, s is the sample standard deviation, n is the sample size, z is the critical value from the standard normal distribution corresponding to the desired confidence level.
Given:
Sample mean X = 2.8 years
Sample standard deviation s = 0.45 years
Sample size n is unknown
Confidence level = 95%
Since we do not know the sample size n, we can use the t-distribution instead of the standard normal distribution to find the critical value. With a 95% confidence level and n-1 degrees of freedom, the critical value is t = 2.093.
Substituting the values, we get:
CI = 2.8 ± 2.093*(0.45/√n)
To find the sample size n, we can solve for it by setting the margin of error to half of the width of the confidence interval, which is equal to 2.093*(0.45/√n):
0.5*(2.093*(0.45/√n)) = 0.025
Simplifying and solving for n, we get:
n ≈ 78
Therefore, a 95% confidence interval for the mean lifetime of all LED TV is:
CI = 2.8 ± 2.093*(0.45/√78) = (2.664, 2.936)
(b) To be 90% confident and have a maximum error of estimation of 0.2, we can use the formula:
n = (z*s/E)^2
where E is the maximum error of estimation and z is the critical value from the standard normal distribution corresponding to the desired confidence level.
Given:
Confidence level = 90%
Maximum error of estimation E = 0.2
Sample standard deviation s = 0.45 years
The critical value corresponding to a 90% confidence level is z = 1.645.
Substituting the values, we get:
n = (1.645*0.45/0.2)^2 ≈ 27.95
Rounding up, we need to add 28 more students to the sample.
(c) To test if the mean GPA is more than 2.5 at a 5% level of significance, we can use a one-tailed t-test with the null and alternative hypotheses:
H0: μ ≤ 2.5
Ha: μ > 2.5
where μ is the population mean GPA.
Given:
Sample mean X = 2.8 years
Sample standard deviation s = 0.45 years
Sample size n is unknown
Level of significance = 5%
We do not know the population standard deviation, so we will use a t-distribution with n-1 degrees of freedom. The test statistic is calculated as:
t = (X - μ) / (s/√n)
To reject the null hypothesis at a 5% level of significance, the t-value must be greater than the critical value from the t-distribution with n-1 degrees of freedom and a one-tailed probability of 0.05. Since the alternative hypothesis is one-tailed, we only need to look up the upper tail of the t-distribution.
The critical value for a one-tailed t-test with 95% confidence and n-1 degrees of freedom is t = 1.729.
Substituting the values, we get:
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CRITICAL THINKING
Solve the problem. Show your work.
Find the total area of the flower and vegetable garden.
Use the formula for the area of a rectangle.
6 ft
4 ft
flowers
vegetables
8 ft
Garden
Answer:
Step-by-step explanation:
Formula of rectangle: base x height (b*h)
base: 8ft
height: 6+4= 10
8 x 10 = 80
Area of rectangle: 80ft
Which equation represents the slope-intercept form of the line below
An equation represents the slope-intercept form of the given line is y= 1/2 x+8. Therefore, option D is the correct answer.
Given that, y-intercept = (0, 8) and slope = 1/2.
The general equation of a straight line is y=mx+c, where m is the gradient, and y = c is the value where the line cuts the y-axis. This number c is called the intercept on the y-axis.
Substitute m=1/2 and c=8 in y=mx+c, we get
y= 1/2 x+8
Therefore, option D is the correct answer.
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Find the exact solutions of the given system equations graphically y= x -3 y=_4x2
The value of exact solutions of the given system equations graphically is,
⇒ x = (1 ± √ 47i) / 8
We have to given that;
Equations are,
y = x - 3
y = 4x²
Now, Substitute the value of y in (ii);
x - 3 = 4x²
4x² - x + 3 = 0
x = - (- 1) ± √(- 1)² - 4×4×3/8
x = (1 ± √1 - 48) / 8
x = (1 ± √- 47) / 8
Thus, The value of exact solutions of the given system equations graphically is,
⇒ x = (1 ± √ 47i) / 8
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Find the value of x, y, and z. The work I did in the problem is how I got it wrong.
In the right triangle, the value of x is 20.78, y is 10.4 and z is 18.
What is the value of x, y, z?
The value of x, y , z is calculated by applying trig ratio as follows;
SOH CAH TOA
SOH = sin θ = opposite /hypothenuse side
TOA = tan θ = opposite side / adjacent side
CAH = cos θ = adjacent side / hypothenuse side
The adjacent side of the right triangle with angle 45 degrees is calculated as;
cos 45 = h/18√2
h = 18√2 x cos (45)
h = 18
h = base of triangle with angle 30⁰;
The value of z is calculated as;
sin 45 = z/18√2
z = 18√2 xsin (45)
z = 18
The value of x is calculated as follows;
cos 30 = 18/x
x = 18/cos30
x = 20.78
The value of y is calculated as follows;
tan 30 = y/18
y = 18 x tan (30)
y = 10.4
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Let M be a 10 × 10 real matrix such that M^2 = M and the
determinant of the
matrix cannot be 1. Does there exist another 10 × 10 real matrix N
such that MN = NM = In ?
(NM)v_i = N(Mv_i) = Nv_i = v_i.
Thus, MN = NM = I, as required.
Yes, such a matrix N exists.
Since M^2 = M, we can write M(M - I) = 0. Therefore, the only possible eigenvalues of M are 0 and 1.
If M has any eigenvalue equal to 0, then the determinant of M is 0, which is not possible according to the problem statement. Hence, all eigenvalues of M are 1.
Since M is a 10 x 10 matrix, it must have a basis of 10 linearly independent eigenvectors corresponding to the eigenvalue 1. Let v1, v2, ..., v10 be such eigenvectors.
Now, we can define N as follows:
For any i and j, if i = j, then Nv_i = v_i.
For any i and j, if i ≠ j, then Nv_i = v_j.
It is easy to verify that this N satisfies MN = NM = I, the identity matrix.
To see this, note that for any i,
(MN)v_i = M(Nv_i) = Mv_i = v_i
and
(NM)v_i = N(Mv_i) = Nv_i = v_i.
Thus, MN = NM = I, as required.
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A store is selling a large selection of men's shirts, and every shirt has the same
price. Kiyoshi buys 6 shirts and gets a $30 discount. His friend Gregory buys 4 shirts
and does not receive a discount. Gregory spends $20 less than Kiyoshi. What is the
price of one shirt without any discount?
O $25
A store is selling a large selection of men's shirts, and every shirt has the same price. Kiyoshi buys 6 shirts and gets a $30 discount. His friend Gregory buys 4 shirts and does not receive a discount. Gregory spends $20 less than Kiyoshi. 25% is the price of one shirt without any discount.
A reduction from the list price of products or services is known as a discount. It denotes the selling of a product for less than its typical cost. In most cases, discounts are expressed as percentages. On the other hand, it could also represent a set discount from the original cost of the goods or services. The difference above the purchase price and the item's par value is the discount.
PERSON SHIRTS COST
Kyoshi 6 6p-30
Gregory 4 4p
DIFFERENCE 20
6p-30-4p=20
6p-4p=20%2B30
2p=50
p=25%
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9. 2 worksheet number #2 in exercies 1 and 2 copy and compelete the table write your anwsers in the simplest form
Fron the sine rule of a right angled triangle, the missing values of table are
Row 1 : [tex]3\sqrt{3}[/tex], 8, 5 ;Row 2 : [tex]11\sqrt{3}[/tex], [tex]8 \sqrt{3}[/tex] ;Row 3 : 22, [tex]6\sqrt{3}[/tex], 10.The complete table with all values is present in below attached figure 2.
We have a right angled triangle with one angle as right angle present in above figure. We have to complete the table present below the figure. The measure of angles of triangle except right angle are 60° and 45°. Also, the side lengths of triangle are 'a', 'b' and 'c' units. Using the sine rule, [tex]\frac{a}{sin(A)} =\frac{ b}{sin (B)} = \frac{c}{sin(C)}[/tex]
Here, A = 30°, B = 60°, C = 90° so, [tex]\frac{a}{sin(30°)} = \frac{ b}{sin(60°)} = \frac{c}{sin(90°)} [/tex]
From the Trigonometry Ratio table of
sin(90°) = 1[tex]sin(60°) = \frac{\sqrt{3}}{2}[/tex][tex]sin(30°) = \frac{1}{2} [/tex]So, [tex] \frac{ a}{ \frac{1}{2} } = \frac{b}{ \frac{ \sqrt{3} }{2} } = \frac{c}{1} [/tex]
[tex]2a= \frac{2b}{ \sqrt{3} } = c [/tex]
Now, consider the first column of table where, a = 11, from equation (1),
[tex]2× 11 = \frac{ 2b}{\sqrt{3}}[/tex]
=> [tex]b = 11\sqrt{3}[/tex] and 2× 11 = c
=> c = 22
Consider the second column of table, where b = 9 then, [tex]a = \frac{2× 9} {\sqrt{3}}[/tex]
=> [tex]a = 2× 3\sqrt{3} = 6\sqrt{3}[/tex]
and [tex] 2a = 2× 6\sqrt{3} = c[/tex]
=> [tex] c = 12\sqrt{3}[/tex]
Consider the third column of table, where c =16 then, 2a = c = 16
=> a = 8
and [tex] c = \frac{2b}{\sqrt{3} }= 16 [/tex]
=> [tex]b = \frac{16\sqrt{3}}{2 } = 8\sqrt{3}[/tex].
Consider the fourth column of table, where [tex]b = 5\sqrt{3}[/tex], then
[tex] 2a = \frac{2× 5\sqrt{3}}{\sqrt{3}} = 10[/tex]
=> a = 5
and [tex] c = \frac{2× 5\sqrt{3}}{\sqrt{3}} = 10[/tex]. Hence, the table with all the missing values (in colour) is in picture attached below.
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Complete question:
The above figure complete question.
9. 2 worksheet number #2 in exercies 1 and 2 copy and compelete the table write your anwsers in the simplest form
sheri walked 15\16 of a mile to school and then 7\8 of a mile to the library. estimate how far sheri walked in total? PLS HELP D:
Answer:
29/16, or 1.81 in decimal form.
Step-by-step explanation:
To solve this, we have to find a common denominator between the 2 fractions.
We cannot simplify 15/16 any further without getting a decimal, so let's change 7/8.
A common denominator between the 2 fractions is 16, so multiply 7/8 by 2 to get 14/16.
Now, we have to find out how far she walked in total, so add 14/16 + 15/16:
29/16
Hope this helps! :)
The marks obtained by the students in physics and in mathematics are as follows. Marks in Physics 35 23 47 17 10 43 9 6 28
Marks in Mathematics 30 33 45 23 8 49 12 4 31
Compute of correlation of ranks.
A. 0.2
B. 0.3
C. 0.7
D. 0.9
The correlation of ranks is approximately 0.2.
Option A is the correct answer.
We have,
To compute the correlation of ranks, we first need to rank the scores in each subject:
Physics: 10, 17, 23, 28, 35, 43, 47
Rank: 1, 2, 3, 4, 5, 6, 7
Mathematics: 4, 8, 12, 23, 30, 31, 33, 45, 49
Rank: 1, 2, 3, 4, 5, 6, 7, 8, 9
Then, we can calculate the differences between the ranks for each student:
Physics ranks: 1-5, 2-3, 3-7, 4-6, 5-1, 6-4, 7-2
Differences: -4, -1, -4, -2, 4, 2, 5
Mathematics ranks: 1-8, 2-6, 3-7, 4-4, 5-1, 6-5, 7-2, 8-3, 9-9
Differences: -7, -4, -4, 0, 4, -1, 5, 5, 0
Next, we can calculate the sum of the products of the differences:
= Sum of products
= (-4)(-7) + (-1)(-4) + (-4)(-4) + (-2)(0) + (4)(4) + (2)(-1) + (5)(5)
= 28 + 4 + 16 + 0 + 16 - 2 + 25
= 87
Finally, we can use the formula for the correlation of ranks:
r = 1 - (6Σd²)/(n(n² - 1))
where d is the difference in ranks and n is the number of scores
Plugging in the values, we get:
r = 1 - (6(87))/(9(81-1))
= 1 - (522)/(648)
= 1 - 0.8056
= 0.1944
= 0.2
Therefore,
The correlation of ranks is approximately 0.2.
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