(b) Prove that if the sequence (4) satisfies lim = L = 0, then a) is unbounded. 71

The confidence level is 1 - α = 1 - 0.005 = 0.995 or approximately 99.5%.

We can use the formula for a confidence interval for a population mean, which is:

[tex]x ± z(α/2) * σ/√n[/tex]

where x is the sample mean, σ is the population standard deviation, n is the sample size, and z(α/2) is the critical value from the standard normal distribution corresponding to the desired confidence level (α).

In this case, the interval is x ± 2.81σ/√n, which is equivalent to z(α/2) = 2.81.

To find the confidence level, we need to solve for α. We can do this by finding the area in the tails of the standard normal distribution that corresponds to z(α/2) = 2.81. Using a standard normal table or a calculator, we find that the area in the right tail is 0.0025, so the area in both tails is 0.005.

Therefore, the confidence level is 1 - α = 1 - 0.005 = 0.995 or approximately 99.5%.

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The minimum sample size required by the economist, if he wants to restrict the margin of error to $480, is 37.

To determine the minimum sample size required to estimate the 99% confidence interval for the average real estate taxes collected by a small town in California with a margin of error of $480, we can use the formula:

[tex]n = \frac{[(z-value)^2 (standard deviation)^2] }{(margin of error)^2}[/tex]

The z-value for a 99% confidence interval is 2.576 (using the z table), and the standard deviation of real estate taxes is $1,330.

Plugging in these values, we get:

[tex]n = \frac{[(2.576)^2 (1,330)^2] }{(480)^2}[/tex]
n = 36.54

Rounding up to the nearest whole number, the minimum sample size required is 37.

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28°49€  arc unit into time is approximately 1.867 hours.

We'll convert the given angles from degrees, minutes, and seconds (or cents and euros as placeholders) to degrees in decimal form. Then, we'll convert the angle from arc units to time units.

a) 39°50¢
To convert 50¢ to degrees, divide by 60 (since 1 degree = 60 minutes):
50¢ / 60 = 0.8333 (rounded to four decimal places)

So, 39°50¢ in decimal form is:
39 + 0.8333 = 39.8333°

b) 42°35¢
To convert 35¢ to degrees:
35¢ / 60 = 0.5833 (rounded to four decimal places)

So, 42°35¢ in decimal form is:
42 + 0.5833 = 42.5833°

c) 15°20€
To convert 20€ to degrees (1 degree = 3600 seconds):
20€ / 3600 = 0.0056 (rounded to four decimal places)

So, 15°20€ in decimal form is:
15 + 0.0056 = 15.0056°

d) 1°59€
To convert 59€ to degrees:
59€ / 3600 = 0.0164 (rounded to four decimal places)

So, 1°59€ in decimal form is:
1 + 0.0164 = 1.0164°

Now, we'll convert 28°49€ from arc units to time units:
28°49€ = 28 + (49 / 3600) = 28.0136° (in decimal form)

To convert degrees to time units, multiply by 24 (since there are 24 hours in a day) and divide by 360 (since there are 360 degrees in a circle):

28.0136° * (24 / 360) = 1.867 (rounded to three decimal places)

So, 28°49€ in time units is approximately 1.867 hours.

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The value of angle ADC is 131⁰.

The value of angle AOC is 262⁰.

The value of angle BAT is 21⁰.

The value of angle ADC is calculated as follows;

The value of the angle at center, AOC = 2 ∠CBA (angle at center is twice angle at circumference)

∠AOC = 2 x 131⁰

∠AOC = 262⁰

The value of arc angle ABC = 262⁰ (intersecting chord theorem)

The value of angle ADC is calculated as;

∠ADC = ¹/₂ arc ABC

∠ADC = ¹/₂ x 262⁰

∠ADC = 131⁰

Angle DAB = 180 - (20 + (131 - 20))

∠DAB = 180 - 131 = 49

The value of angle BAT is calculated as;

∠BAT = 90 - (20 + 49)

∠BAT = 21⁰

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Answer:

y = -3(x - 5)^2 + 3

Step-by-step explanation:

Because we're given the maximum/vertex of the quadratic function and at least one of the roots, we can find the equation of the quadratic equation using the vertex form which is

[tex]y = a(x-h)^2+k[/tex], where a is a constant (determine whether parabola will have maximum or minimum), (h, k) is the vertex (a maximum for this problem), and (x, y) are any point on the parabola:

Since our maximum/vertex is (5, 3), and one of our roots is (6, 0), we can plug everything in and solve for a:

[tex]0=a(6-5)^2+3\\0=a(1)^2+3\\0=a+3\\-3=a[/tex]

Thus, the general equation (without distribution) is y = -3(x - 5)^2 + 3

Answer: 26

Step-by-step explanation:

So what its basically asking is for you to find the hypotenuse because you can see that the rectangle splits in half with the green line.

So to find the hypotenuse you would use these steps:

1. formula for hypotenuse  

[tex]\sqrt{a^2+b^2}[/tex]

2. plug in numbers

[tex]\sqrt{10^2+24^2}=26[/tex]

A p-value less than 0.05 indicates that there is a significant difference between the average expected incomes of the different ethnic groups.

We will perform an ANOVA (Analysis of Variance) test to determine if there are any significant differences between the average expected incomes of the 5 ethnic groups.

a. Compute the Chi Square Statistic: The Chi Square Statistic is not applicable in this case, as it is used for categorical data. Instead, we will use the F Statistic for ANOVA.

b. Degrees of freedom: For ANOVA, there are two degrees of freedom to calculate: - df_between (numerator): This is the degrees of freedom between groups, which is equal to the number of groups minus 1. There are 5 ethnic groups, so df_between = 5 - 1 = 4. - df_within (denominator): This is the degrees of freedom within groups, which is equal to the total number of observations minus the number of groups. The total number of observations is 922, so df_within = 922 - 5 = 917.

c. Compute the F Statistic: Use an ANOVA calculator or software to input the data and calculate the F Statistic.

d. Degrees of freedom in the numerator and denominator: - Numerator degrees of freedom: 4 (as calculated in part b) - Denominator degrees of freedom: 917 (as calculated in part b) After computing the F Statistic, use an online calculator or software to find the p-value.

A p-value less than 0.05 indicates that there is a significant difference between the average expected incomes of the different ethnic groups.

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The probability that a randomly chosen voter from the survey  is under 50 years old is 0.75

From the question, we have the following parameters that can be used in our computation:

The table of values

Where we have

Voters under 50 years old = 847 + 804 + 773

Total = 3228

So, the required probability is

P = (847 + 804 + 773)/3228

Evaluate

P = 0.75

Hence, the probability is 0.75

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[tex]\frac{(λ-b)}{a}[/tex] is an unbiased estimator for λ

a) To show that λ + X is an unbiased estimator for λ, we need to show that its expected value is equal to λ.

We know that λ is an unbiased estimator for λ, which means that E(λ) = λ.

Now, let's calculate the expected value of λ + X:

E(λ + X) = E(λ) + E(X)

Since E(X) = 0 (given that X has mean zero), we have:

E(λ + X) = E(λ) + 0

E(λ + X) = E(λ)

E(λ + X) = λ

Therefore, λ + X is an unbiased estimator for λ.

b) Given E(λ) = aλ + b and a ≠ 0, we can find an unbiased estimator for λ by solving for λ in terms of [tex]\frac{(λ-b)}{a}[/tex].

We have:

E(λ) = aλ + b

Dividing both sides by a, we get:

[tex]\frac{E(λ)}{a} = λ +\frac{b}{a}[/tex]

Subtracting [tex]\frac{b}{a}[/tex] from both sides, we have:

[tex]\frac{E(V)}{a} - \frac{b}{a} } =  λ[/tex]

Simplifying, we get:

[tex]\frac{(λ - b)}{a} = \frac{E(λ - b)}{a} - \frac{b}{a}[/tex]

Therefore, [tex]\frac{(λ-b)}{a}[/tex] is an unbiased estimator for λ.

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The length of one side of the square is 3, as solving for "x" in the equation A(x) = x² - 6x + 9 yields (x - 3)² = 0, which has a solution of x = 3.

The area of a square is typically calculated using the formula A = s², where "s" represents the length of one side of the square.

In this problem, we are given the area of the square as A(x) = x² - 6x + 9.

To find the length of one side of the square, we need to solve for "x" in the equation A(x) = x² - 6x + 9.

Setting A(x) equal to zero

x² - 6x + 9 = 0

Factoring the quadratic

(x - 3)² = 0

Expanding the squared term

x - 3 = 0

Solving for "x"

x = 3

Therefore, the length of one side of the square is 3.

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Hawthorne Effect

You are conducting a research study and giving a group of participants an accelerometer. When analyzing the data, you notice that all participants had the highest levels of physical activity on Day 1. You decide to exclude this data. Excluding this data is an example of trying to avoid: Hawthorne Effect

The Hawthorne Effect refers to the phenomenon where participants modify their behavior in response to being observed or aware of being part of a study. By excluding the data from Day 1, you are trying to avoid the potential influence of this effect on the study results.

You collect your data by watching the employees during their work breaks. If employees are aware that you are observing them, this can affect your study's results. For example, you may record higher or lower smoking rates than are genuinely representative of the population under study.

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This is because as the level of confidence decreases, the corresponding z-score becomes smaller, which in turn results in a smaller margin of error.

(a) To find a 90% confidence interval for the proportion p, we can use the formula:

CI = p ± z*(sqrt(p*(1-p)/n))

where p is the sample proportion, n is the sample size, and z is the z-score corresponding to the desired level of confidence (in this case, 90%).

We are given that p = 330/550 = 0.6 and n = 550. Using a standard normal distribution table, the z-score for a 90% confidence interval is approximately 1.645.

Substituting these values into the formula, we get:

CI = 0.6 ± 1.645*(sqrt(0.6*(1-0.6)/550))

= 0.6 ± 0.042

= (0.558, 0.642)

Therefore, a 90% confidence interval for the proportion of voters who indicated their preference for the candidate is 0.558 to 0.642.

(b) The margin of error for this 90% confidence interval is given by:

ME = z*(sqrt(p*(1-p)/n))

where z is the z-score corresponding to the desired level of confidence and p and n are as before.

Substituting the values we obtained earlier, we get:

ME = 1.645*(sqrt(0.6*(1-0.6)/550))

= 0.042

Therefore, the margin of error for this 90% confidence interval is 0.042.

(c) Without doing any calculations, we can see that the margin of error for an 80% confidence interval will be smaller than that for a 90% confidence interval. This is because as the level of confidence decreases, the corresponding z-score becomes smaller, which in turn results in a smaller margin of error.

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The conditions that must hold for inferential procedures to be valid for this scenario are:

(b) the two sample groups must be independent of each other, (c) the data distribution for both men's and women's hemoglobin levels must be normally distributed or each sample size must be larger than 30, and (d) the observations within each sample must be independent.

(b) The two sample groups must be independent of each other because the samples should not be related in any way that could influence the results.

(c) The data distribution for both men's and women's hemoglobin levels must be normally distributed or each sample size must be larger than 30. If the data is not normally distributed, then the Central Limit Theorem can be applied if the sample size is greater than 30.

(d) The observations within each sample must be independent to avoid bias in the results. This means that each observation should not be influenced by any other observation.

(a) The expected count for each level of the categorical variable must be at least 5. This condition is relevant for Chi-square tests of independence, which are not being used in this scenario.

(e) There must be 10 successes and 10 failures in each sample. This condition is relevant for tests of proportions, which are not being used in this scenario.

So b,c and d are correct options.

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(a) The values of u is 140 g and o is 13.42 g. (b) The value of K in (300 - K. 300 + K) grams is 27.15 g. C) The weights of the middle 96.6% of fruit cups are between (L1, L2) grams. The values of LI is 272.85 g and L2 is 327.15 g.

(a) The mean weight of blueberries is:

300 g - 160 g = 140 g

The standard deviation of the weight is:

Var(X + Y) = Var(X) + Var(Y)

Adding the variances:

15^2 = 10^2 + o^2

Solving for o:

o = sqrt(15^2 - 10^2) = 13.42 g

Therefore, the values of u and o are u = 140 g and o = 13.42 g.

(b) Since the distribution is normal, we can use the standard normal distribution to find K.

The middle 96.6% of a standard normal distribution corresponds to the interval (-1.81, 1.81) (using a table or calculator). Therefore,

K = 1.81 * 15 = 27.15 g

Therefore, the weights of the middle 96.6% of fruit cups are between 300 - 27.15 = 272.85 g and 300 + 27.15 = 327.15 g.

(c) Using the standard normal distribution to find the corresponding interval on the standard normal scale:

(-1.81, 1.81)

We can then scale this interval to the distribution of the weight of fruit cups by dividing by the standard deviation and multiplying by 15 g:

L1 = 300 + (-1.81) * 15 = 272.85 g

L2 = 300 + 1.81 * 15 = 327.15 g

Therefore, the weights of the middle 96.6% of fruit cups are between 272.85 g and 327.15 g.

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The total cost of tuition, books, and fees for this student when attending college for four years will be $64651

Assuming that the annual increase of $750 per year is compounded each year

we can use the formula for the future value of an annuity to calculate the total cost of tuition, books, and fees for the four years:

[tex]FV = PMT \frac{((1 + r)^n - 1)}{r}[/tex]

In this case, PMT = $15,000, r = 750/15000 = 0.05, and n = 4.

Plugging in these values, we get:

Total cost or FV = $15,000 x ((1 + 0.05)⁴ - 1) / 0.05

FV = $15,000 x(1.2155)-1)/0.05

FV = $15,000 x 0.2155/0.05

FV = $15,000 x4.31

FV = $64651

Hence, the total cost of tuition, books, and fees for this student when attending college for four years will be $64651

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Answer:

replace

y

with

0

and solve for

x

.

Step-by-step explanation:

x=-1/3,2

The plot of annual sheep livestock numbers against the year in Asia from 1961 to 2007 would provide a visual representation of the trends, fluctuations, peaks and valleys, patterns, and outliers in sheep population, offering valuable insights into the dynamics of sheep farming in Asia during the period.

The plot would show a graph with the years on the x-axis and the annual sheep livestock numbers on the y-axis. The plot would display data points connected by lines, representing the annual sheep livestock numbers for each year from 1961 to 2007.

The main features of the plot may include the following:

Trend: The plot would show the overall trend of sheep livestock numbers in Asia from 1961 to 2007. It may reveal whether the sheep population has increased, decreased, or remained relatively stable over time.

Fluctuations: The plot may show fluctuations or variations in sheep livestock numbers from year to year. These fluctuations could be due to various factors such as changes in farming practices, climate conditions, disease outbreaks, or economic factors.

Peaks and Valleys: The plot may display peaks and valleys, indicating the highest and lowest points of annual sheep livestock numbers during the period. These peaks and valleys may provide insights into significant events or trends affecting sheep population in Asia.

Patterns: The plot may reveal patterns or cycles in sheep livestock numbers over time. For example, there may be recurring patterns of increase or decrease in sheep population at regular intervals or irregular patterns that indicate changes in sheep farming practices or market demand.

Outliers: The plot may also show outliers, which are data points that deviate significantly from the overall trend. These outliers could represent exceptional years with unusually high or low sheep livestock numbers, which may warrant further investigation.

Therefore, the plot of annual sheep livestock numbers against the year in Asia from 1961 to 2007 would provide a visual representation of the trends, fluctuations, peaks and valleys, patterns, and outliers in sheep population, offering valuable insights into the dynamics of sheep farming in Asia during the period.

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A)

1) the area of the shapes are given as follows:

2) The probabilities are:

B)

1) The area of Board is = 384

The area of the circles are:

2) The probabilities are:

A)

1)

2)

i) The probability of coin landing in the circle   is given by   the ratio of the area of the circle to the area of the   board:

P(circle) = Area of circle / Area of board = 36pi / 450 ≈ 0.2513

ii)

The probability of a coin landing in the triangle but not in the circle is

P(triangle and not circle) = (A are of triangle -  area of circle) / Area of board = (225 - 36pi) / 450 ≈ 0.2376

iii) The probability of a coin landing in neither the circle nor the triangle is  P (neither) = 1 - (P(circle) +  P (triangle and not circle)) = 1 - (0.2513 + 0.2376 ) = 0.5111

B)

1)

2)

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The mean of the data set is 9.33 minutes.

To find mean of the data set, we will add all the values and divide by the total number of values.

In this case, the sum of the values is:

= 0 + 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 + 22

= 112

There are 12 values in the data set, so the mean is:

= Sum of values / Total number of values

= 112 / 12

= 9.3333

= 9.33 minutes.

Therefore, the mean of the data set is 9.33 minutes.

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a) P(30 ≤ x ≤ 32) = 0.3446.

b) P(30 ≤ x ≤ 32) = 0.2868.

c) The probability of getting values between 30 and 32 for x is higher in part (b) than in part (a).

We have,

(a)

The mean of the distribution is = 30 and the standard deviation is = 28.

For a sample size n = 31, the sample mean x follows a normal distribution with mean = 30 and standard deviation = /√n = 28/√31 = 5.02 (approx.).

Therefore, x ~ N(30, 5.02).

The probability P(30 ≤ x ≤ 32) can be found by standardizing the values using the formula z = (x - ) / , where z is the standard normal variable.

z1 = (30 - 30) / 5.02 = 0

z2 = (32 - 30) / 5.02 = 0.40

P(30 ≤ x ≤ 32) = P(0 ≤ z ≤ 0.40) = 0.3446 (approx.)

Therefore, x = 30, x = 5.02, and P(30 ≤ x ≤ 32) = 0.3446 (approx.).

(b)

For a sample size n = 62, the sample mean x follows a normal distribution with mean = 30 and standard deviation = /√n = 28/√62 = 3.56 (approx.).

Therefore, x ~ N(30, 3.56).

The probability P(30 ≤ x ≤ 32) can be found using the same method as in part (a).

z1 = (30 - 30) / 3.56 = 0

z2 = (32 - 30) / 3.56 = 0.56

P(30 ≤ x ≤ 32) = P(0 ≤ z ≤ 0.56) = 0.2868 (approx.)

Therefore, x = 30, x = 3.56, and P(30 ≤ x ≤ 32) = 0.2868 (approx.).

(c)

The standard deviation of part (b) is smaller than part (a) because of the larger sample size.

Therefore, the distribution about x is narrower in part (b) than in part (a). This means that the sample mean x in part (b) is likely to be closer to the population mean than the sample mean x in part (a).

As a result, the probability of getting values between 30 and 32 for x is higher in part (b) than in part (a).

Thus,

a) P(30 ≤ x ≤ 32) = 0.3446.

b) P(30 ≤ x ≤ 32) = 0.2868.

c) The probability of getting values between 30 and 32 for x is higher in part (b) than in part (a).

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We have found the measures of CD, CE, HD, GD, HG, and HF in triangle BCD, given that H is the circumcenter of the triangle.

In triangle BCD, the circumcenter H is the point where the perpendicular bisectors of the sides of the triangle intersect. This point is equidistant from the three vertices of the triangle.

Using the properties of the circumcenter, we can find the measures of various sides and angles of the triangle:

CD = 2FD, where FD is the foot of the perpendicular from H to CD.

CE = BE = 26, since H is equidistant from B and C.

HD = HC = 33, since H is equidistant from D and C.

GD = 1/2BD = 1/2(58) = 29, since H is equidistant from B and D.

HG = √HD² - GD² = √33² - 29² = 2√62 ≈ 15.75, using the Pythagorean theorem.

HF = √HD² - FD² = √33² - 32² = √65 ≈ 8.06, using the Pythagorean theorem.

Therefore, we have found the measures of CD, CE, HD, GD, HG, and HF in triangle BCD, given that H is the circumcenter of the triangle.

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The matrix of constants you provided [[12],[11],[4]] represents a system of linear equations with 3 variables and 1 equation, which is an underdetermined system.

To write the system of equations in the form Ax = b, where A is the matrix of coefficients, x is the vector of variables, and b is the vector of constants, we can set:

12x1 = 11x2 + 4x3

where x1, x2, and x3 are the variables of the system.

Note that this system has infinitely many solutions, since there are more variables than equations. In other words, we can choose any value for one of the variables and solve for the other two.

Assuming that you meant to ask for the simplified form of the expression:

The formula is (a × x) = (a × x) + (b × a × x)

We can first simplify the expression by combining like terms:

(1 + 4b) is equal to (axe) + b(ax) + c(a × x) + d(a × x)√(ax)

Therefore, the simplified expression is:

(1 + 4b)√()axe) + b(ax) + c(a × x) + d(a × x)√(ax)

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The point estimate for the difference between the two population means is $1.25.

To find the point estimate for the difference between the two population means, subtract the sample mean of company B from the sample mean of company A:

Point estimate = $17.75 - $16.50 = $1.25

This means that the average hourly wage in company A is estimated to be $1.25 higher than the average hourly wage in company B. It's important to note that this is just a point estimate and not a conclusive result. To determine if this difference is statistically significant, further hypothesis testing would be needed.

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(i) c = 1/2 and the joint pdf is f(x, y) = (x+y)/2 over the unit square.

(ii) 1/16

(iii) 1/9

iv)  5/3

(v) E(X+Y) = 5/6.

we have,

i.

In order for f(x, y) to be a valid joint pdf, it must satisfy two conditions:

It must be non-negative for all (x,y)

The integral over the entire support must equal 1.

To satisfy the first condition, we need c(x+y) to be non-negative.

This is true as long as c is non-negative and x+y is non-negative over the support, which is the unit square [0,1]x[0,1]. Since x and y are both non-negative over the unit square, we need c to be non-negative as well.

To satisfy the second condition, we integrate f(x, y) over the unit square and set it equal to 1:

1 = ∫∫ f(x, y) dx dy

 = ∫∫ c(x+y) dx dy

 = c ∫∫ (x+y) dx dy

 = c [∫∫ x dx dy + ∫∫ y dx dy]

 = c [∫ 0^1 ∫ 0^1 x dx dy + ∫ 0^1 ∫ 0^1 y dx dy]

 = c [(1/2) + (1/2)]

 = c

ii.

P(X < 0.5, Y < 0.5) can be found by integrating the joint pdf over the region where X < 0.5 and Y < 0.5:

P(X < 0.5, Y < 0.5) = ∫ 0^0.5 ∫ 0^0.5 (x+y)/2 dy dx

                    = ∫ 0^0.5 [(xy/2) + (y^2/4)]_0^0.5 dx

                    = ∫ 0^0.5 [(x/4) + (1/16)] dx

                    = [(x^2/8) + (x/16)]_0^0.5

                    = (1/32) + (1/32)

                    = 1/16

iii.

P(Y<X) can be found by integrating the joint pdf over the region where

Y < X:

P(Y < X) = ∫ 0^1 ∫ 0^x (x+y)/2 dy dx

        = ∫ 0^1 [(xy/2) + (y^2/4)]_0^x dx

        = ∫ 0^1 [(x^3/6) + (x^3/12)] dx

        = (1/9)

iv.

P(X+Y) < 5 can be found by integrating the joint pdf over the region where X+Y < 5:

P(X+Y < 5) = ∫ 0^1 ∫ 0^(5-x) (x+y)/2 dy dx

          = ∫ 0^1 [(xy/2) + (y^2/4)]_0^(5-x) dx

          = ∫ 0^1 [(x(5-x)/2) + ((5-x)^2/8)] dx

          = 5/3

v.

The expected value of XY can be found by integrating the product xy times the joint pdf over the entire support:

E(XY) = ∫∫ xy f(x, y) dx dy

E(XY) = ∫∫ xy (x+y)/2 dx dy

= ∫∫ (x^2y + xy^2)/2 dx dy

= ∫ 0^1 ∫ 0^1 (x^2y + xy^2)/2 dx dy

= ∫ 0^1 [(x^3*y/3) + (xy^3/6)]_0^1 dy

= ∫ 0^1 [(y/3) + (y/6)] dy

= 1/4

The expected value of X+Y can be found by integrating the sum (x+y) times the joint pdf over the entire support:

E(X+Y) = ∫∫ (x+y) f(x, y) dx dy

      = ∫∫ (x+y) (x+y)/2 dx dy

      = ∫∫ [(x^2+2xy+y^2)/2] dx dy

      = ∫ 0^1 ∫ 0^1 [(x^2+2xy+y^2)/2] dx dy

      = ∫ 0^1 [(x^3/3) + (xy^2/2) + (y^3/3)]_0^1 dy

      = ∫ 0^1 [(1/3) + (y/2) + (y^2/3)] dy

      = 5/6

Thus,

(i) c = 1/2 and the joint pdf is f(x, y) = (x+y)/2 over the unit square.

(ii) 1/16

(iii) 1/9

iv)  5/3

(v) E(X+Y) = 5/6.

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Its Fourier transform is also 0.

The Fourier transform of a function f(t) is defined as:

F(w) = (1/√(2π)) ∫[from -∞ to +∞] f(t) e^(-iwt) dt

Let's find the Fourier transform of the given function f(t) = sin 3t, k≤|t|≤2k and 0, |t|>2k.

For k≤|t|≤2k, we can write:

f(t) = sin 3t

= (1/2i) (e^(i3t) - e^(-i3t))

Using the Fourier transform properties, we can write:

F(w) = (1/2i) [∫[from -2k to -k] e^(i3t) e^(-iwt) dt + ∫[from k to 2k] e^(i3t) e^(-iwt) dt]

Applying the integral formula ∫ e^(ax) dx = (1/a) e^(ax) + C, we get:

F(w) = (1/2i) [(1/i(3-w))(e^(i(3-w)2k) - e^(i(3-w)k)) + (1/i(3+w))(e^(i(3+w)k) - e^(i(3+w)2k))]

Simplifying the above expression, we get:

F(w) = (1/2) [(sin(2kw-3) - sin(kw-3))/(kw-3) + (sin(kw+3) - sin(2kw+3))/(kw+3)]

For |t|>2k, f(t) = 0. Thus, its Fourier transform is also 0.

Therefore, the correct option is d. none of the above.

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The list of the temperature from coldest to warmest temperature includes:

The temperatures (Fahrenheit) of the 4 cities from coldest to warmest includes -9 degrees, -7 degrees, -6 degrees, 5 degrees and 12 degrees.

The coldest temperature is -9 degrees, followed by -7 degrees, then -6 degrees. The positive temperature is  5 degrees and 12 degrees.

We must note these temperatures are in Fahrenheit which is not the standard unit of measurement used in all countries, so, we must specify the unit when reporting temperatures.

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Answer:

Speed= distance/ time

Speed= (2/3)/(3/4) = 0.88 miles/hr