A factory's worker productivity is normally distributed. One worker produces an average of 76 unita per day with a standard deviation of 23 . Another worker produces ot an average rate of 65 units per day with a standard deviation of 22. A. What is the probability that in a single day worker 1 will outproduce worker 2 ? Probabily = B. What is the probability that during one week ( 5 working dayo), worker 1 will outproduce worker 2 ? Probability =

The probability that at least 5 flights are on time is 0.3676.

The probability that at most 3 flights are on time is 0.0081.

The probability that exactly 3 flights are on time is 0.2668.

We have,

To solve these probability questions, we can use the binomial distribution.

The binomial distribution is appropriate here because we have a fixed number of independent trials (7 flights) with two possible outcomes (on time or not on time) and a known probability of success (90% or 0.9).

The probability that at least 5 flights are on time can be calculated by summing the probabilities of having 5, 6, or 7 flights on time:

P(at least 5 flights on time) = P(5 flights on time) + P(6 flights on time) + P(7 flights on time)

[tex]= (^7C_5) (0.9^5) (0.1^2) + (^7C_6) (0.9^6) (0.1^1) + (^7C_7) (0.9^7) (0.1^0)[/tex]

= 0.3676

The probability that at most 3 flights are on time can be calculated by summing the probabilities of having 0, 1, 2, or 3 flights on time:

P(at most 3 flights on time) = P(0 flights on time) + P(1 flight on time) + P(2 flights on time) + P(3 flights on time)

[tex]= (^7C_ 0)(0.9^0) (0.1^7) + (^7 C_ 1) (0.9^1) (0.1^6) + (^7C_ 2) (0.9^2) (0.1^5) + (^7C_ 3) (0.9^3) (0.1^4)[/tex]

= 0.0081

The probability that exactly 3 flights are on time can be calculated using the binomial formula:

P(exactly 3 flights on time)

[tex]= (^7C_ 3) (0.9^3) (0.1^4)[/tex]

= 0.2668

Therefore,

The probability that at least 5 flights are on time is 0.3676.

The probability that at most 3 flights are on time is 0.0081.

The probability that exactly 3 flights are on time is 0.2668.

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The standard deviation of the finishing times in the 200-meter dash population is approximately 2.65 seconds.

To find the standard deviation of a population, we can use the following formula:

σ = √(Σ(x - μ)² / N)

Where:

σ represents the standard deviation of the population.

Σ denotes the summation symbol, which means to sum up the values.

x represents each individual value in the population.

μ represents the mean (average) of the population.

N represents the total number of values in the population.

Let's calculate the standard deviation for the given finishing times of the 200-meter dash:

Finishing times: 32, 25, 29, 26, 25, 25

Step 1: Calculate the mean (μ)

μ = (32 + 25 + 29 + 26 + 25 + 25) / 6

= 162 / 6

= 27

Step 2: Calculate the squared differences from the mean (x - μ)² for each value:

(32 - 27)² = 25

(25 - 27)² = 4

(29 - 27)² = 4

(26 - 27)² = 1

(25 - 27)² = 4

(25 - 27)² = 4

Step 3: Sum up the squared differences:

Σ(x - μ)² = 25 + 4 + 4 + 1 + 4 + 4 = 42

Step 4: Calculate the standard deviation (σ):

σ = √(Σ(x - μ)² / N)

= √(42 / 6)

= √7

≈ 2.65 (rounded to two decimal places)

Therefore, the standard deviation of the population is approximately 2.65 seconds.

The standard deviation measures the spread or variability of the data in a population. It indicates how much the individual values deviate from the mean.

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In summary, the inverse Laplace transforms of the given expressions are:

3.1: -e^(-3t/2) + 4e^(-t)

3.2: cos(2t) - sin(2t)

3.3: e^(5 + √19)t + e^(5 - √19)t

To determine the inverse Laplace transform in its simplest form, we need to find the function in the time domain corresponding to the given Laplace transform expression. In the first case, the Laplace transform is 3/(2s + 3)(s + 2s + 2). In the second case, the Laplace transform is (3 - s)/(s² + 4). In the third case, the Laplace transform is 1/(8 - 12s + s² + s + s - 2). The second paragraph will provide a step-by-step explanation of finding the inverse Laplace transform for each case.

3.1: To find the inverse Laplace transform of 3/(2s + 3)(s + 2s + 2), we first factorize the denominator as (2s + 3)(s + 1). Then, using partial fraction decomposition, we express the given expression as A/(2s + 3) + B/(s + 1), where A and B are constants. Solving for A and B, we get A = -1 and B = 4. Therefore, the inverse Laplace transform of 3/(2s + 3)(s + 2s + 2) is -e^(-3t/2) + 4e^(-t).

3.2: The Laplace transform expression (3 - s)/(s² + 4) can be simplified by completing the square in the denominator. After completing the square, we get (s - 0)² + 4, which is in the form of a shifted complex number. Therefore, we can use the inverse Laplace transform property to find the time-domain function. The inverse Laplace transform of (3 - s)/(s² + 4) is e^(0t)cos(2t) - e^(0t)sin(2t), which simplifies to cos(2t) - sin(2t).

3.3: For the expression 1/(8 - 12s + s² + s + s - 2), we combine like terms to obtain 1/(s² - 10s + 6). Using the quadratic formula, we find the roots of the denominator as s = 5 ± √19. Applying partial fraction decomposition, we write the expression as A/(s - (5 + √19)) + B/(s - (5 - √19)), where A and B are constants. After finding the values of A and B, we substitute the inverse Laplace transform of each term, resulting in e^(5 + √19)t + e^(5 - √19)t.

In summary, the inverse Laplace transforms of the given expressions are:

3.1: -e^(-3t/2) + 4e^(-t)

3.2: cos(2t) - sin(2t)

3.3: e^(5 + √19)t + e^(5 - √19)t

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The probability density function of a continuous random variable X 3x²/8

P(0 ≤ X ≤ 1) = 1/8

E(X) = 3/2

E(X²) = 12/5

Var(X) = 3/20

σ(X) ≈ 0.3464

The values for the given probability density function (pdf), we can use the properties of continuous random variables.

P(0 ≤ X ≤ 1):

This probability, we need to integrate the pdf over the range [0, 1]:

P(0 ≤ X ≤ 1) = ∫[0,1] f(x) dx

Integrating the pdf f(x) = 3x²/8 over the range [0, 1]:

P(0 ≤ X ≤ 1) = ∫[0,1] 3x²/8 dx

Integrating 3x²/8, we get:

P(0 ≤ X ≤ 1) = [x³/8] evaluated from 0 to 1

P(0 ≤ X ≤ 1) = (1³/8) - (0³/8)

P(0 ≤ X ≤ 1) = 1/8

Therefore, P(0 ≤ X ≤ 1) = 1/8.

E(X) - Expected Value of X:

The expected value, we need to calculate the mean of the pdf:

E(X) = ∫[0,2] x × f(x) dx

Substituting the pdf f(x) = 3x²/8:

E(X) = ∫[0,2] x × (3x²/8) dx

E(X) = ∫[0,2] (3x³/8) dx

E(X) = [3x⁴/32] evaluated from 0 to 2

E(X) = (3 × 2⁴/32) - (3 × 0⁴/32)

E(X) = 48/32

E(X) = 3/2

Therefore, E(X) = 3/2.

E(X²) - Expected Value of X²:

To find the expected value of X², we calculate the mean of X²:

E(X²) = ∫[0,2] x² × f(x) dx

Substituting the pdf f(x) = 3x²/8:

E(X²) = ∫[0,2] x² × (3x²/8) dx

E(X²) = ∫[0,2] (3x⁴/8) dx

E(X²) = [3x⁵/40] evaluated from 0 to 2

E(X²) = (3 × 2⁵/40) - (3 × 0⁵/40)

E(X²) = 96/40

E(X²) = 12/5

Therefore, E(X²) = 12/5.

Var(X) - Variance of X:

The variance is calculated as the difference between the expected value of X² and the square of the expected value of X:

Var(X) = E(X²) - (E(X))²

Substituting the values we calculated:

Var(X) = 12/5 - (3/2)²

Var(X) = 12/5 - 9/4

Var(X) = (48 - 45)/20

Var(X) = 3/20

Therefore, Var(X) = 3/20.

σ(X) - Standard Deviation of X:

The standard deviation is the square root of the variance:

σ(X) = √(Var(X))

σ(X) = √(3/20)

σ(X) = √(3)/√(20)

Simplifying the square root:

σ(X) = √(3)/√(4 × 5)

σ(X) = √(3)/2√5

Therefore, σ(X) = √(3)/2√5 (rounded to 4 decimal places).

To summarize the results:

P(0 ≤ X ≤ 1) = 1/8

E(X) = 3/2

E(X²) = 12/5

Var(X) = 3/20

σ(X) ≈ 0.3464

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The spherical coordinates of the point (1, 7/2, 1) in cylindrical coordinates are (ρ, θ, φ) = (3/2, arctan(2), arccos(1/√6)).

To convert the point (1, 7/2, 1) from cylindrical coordinates to spherical coordinates, we need to find the values of ρ, θ, and φ.

In cylindrical coordinates, the point is represented as (ρ, θ, z), where ρ is the radial distance from the z-axis, θ is the azimuthal angle measured from the positive x-axis, and z is the height.

Given that ρ = 1, θ is not provided, and z = 1, we can find the values of ρ, θ, and φ as follows:

1. Radial distance (ρ):

  ρ is the distance from the origin to the point in the xy-plane. In this case, ρ = 1.

2. Azimuthal angle (θ):

  The angle θ is measured from the positive x-axis in the xy-plane. Since θ is not provided, we cannot determine its value.

3. Polar angle (φ):

  The angle φ is measured from the positive z-axis. To find φ, we can use the equation φ = arccos(z/√(ρ² + z²)). Substituting the given values, φ = arccos(1/√(1² + 1²)) = arccos(1/√2) = arccos(1/√6).

Therefore, the spherical coordinates of the point (1, 7/2, 1) in cylindrical coordinates are (ρ, θ, φ) = (1, θ, arccos(1/√6)).

Note: The value of θ cannot be determined with the given information.

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The z-value for P(Z>c) is 0.99. The percentile is 15.9%. The probability is 0.9963. The test score that is 0.8 standard deviations above the mean is 121.

1. For a standard normal distribution, find the boundary c where:

P(Z < c) = 6.89%

For a standard normal distribution, the z-value for P(Z < c) = 6.89% can be calculated as:

z = invNorm(0.0689) ≈ -1.49

We know that the standard normal distribution is symmetric about 0.

Therefore, we can flip the inequality and say:P(Z > c) = 1 - P(Z < c) = 1 - 0.0689 = 0.9311 = 93.11%

Thus, the z-value for P(Z > c) = 83.18% can be calculated as:

z = invNorm(0.8318) ≈ 0.99

2. To find the percentile associated with a z-value, we can use the standard normal distribution table.

For z = -1, the area to the left of z is 0.1587. This means that the percentile associated with z = -1 is 15.87%.

Therefore, the answer is 15.9% (rounded to the nearest tenth of a percent).

3. Here, the mean (μ) = 2.9 years and the standard deviation (σ) = 0.7 years. We are asked to find the probability that the battery will last longer than 1 year. We can find this probability by standardizing the variable x (which represents the battery life in years) as follows:

z = (x - μ) / σz = (1 - 2.9) / 0.7 ≈ -2.71

Now we can use a standard normal distribution table (or calculator) to find P(Z > -2.71).

This probability is approximately 0.9963. Therefore, the probability that the battery will last longer than 1 year is 99.6% (rounded to one decimal place)

4. Here, the mean (μ) = 56.9 inches and the standard deviation (σ) = 1.4 inches. We are asked to find the probability that a randomly chosen child has a height between 54.5 and 54.7 inches. We can find this probability by standardizing the variable x (which represents the height in inches) as follows:

z1 = (54.5 - 56.9) / 1.4 ≈ -1.71z2 = (54.7 - 56.9) / 1.4 ≈ -1.57

Now we can use a standard normal distribution table (or calculator) to find P(-1.71 < Z < -1.57). This probability is approximately 0.0370.

Therefore, the probability that the height of a randomly chosen child is between 54.5 and 54.7 inches is 3.7% (rounded to one decimal place).

5. We are given that the mean (μ) = 6.3 inches and the standard deviation (σ) = 1 inch. We know that 4.3% of the population is outside the range of (μ - 1.5σ) to (μ + 1.5σ). That is:

P(Z < -1.5) + P(Z > 1.5) = 0.043

We can use a standard normal distribution table (or calculator) to find that:

P(Z < -1.5) = 0.0668P(Z > 1.5) = 0.0668

Therefore, the range of head breadths that will fit all men except those with head breadths that are in the smallest 4.3% or largest 4.3% is (μ - 1.5σ) to (μ + 1.5σ).

We can calculate this range as follows:

Lower bound: μ - 1.5σ = 6.3 - 1.5(1) = 4.8 inches

Upper bound: μ + 1.5σ = 6.3 + 1.5(1) = 7.8 inches

Therefore, the minimum head breadth that will fit the clientele is 4.8 inches, and the maximum head breadth that will fit the clientele is 7.8 inches (both rounded to one decimal place

6. Here, the mean (μ) = 105 and the standard deviation (σ) = 20. We are asked to find the test score that is 0.8 standard deviations above the mean. We can use the formula for standardizing a variable x (which represents the test score) to a z-value as follows:

z = (x - μ) / σ = 0.8

standard deviations above the mean is equivalent to a z-value of 0.8.

Therefore, we can plug in z = 0.8, μ = 105, and σ = 20 into the formula and solve for x:

x = zσ + μx = 0.8(20) + 105x = 16 + 105x = 121

Therefore, the test score that is 0.8 standard deviations above the mean is 121.

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Let's denote the amount of the 70% HCl solution to be mixed as x ml. In the 85 ml of a 10% HCl solution, we have 0.10 * 85 = 8.5 ml of HCl. In x ml of the 70% HCl solution, we have 0.70x ml of HCl.

When the two solutions are mixed, the total volume of the resulting solution will be 85 + x ml. To create a 20% HCl solution, we want the amount of HCl in the mixture to be 20% of the total volume. Therefore, we can set up the equation: 0.70x + 8.5 = 0.20 * (85 + x). Simplifying the equation, we have: 0.70x + 8.5 = 17 + 0.20x; 0.50x = 8.5 - 17; 0.50x = -8.5; x = -8.5 / 0.5; x = -17.

Since the amount of solution cannot be negative, there is no solution for this problem. It is not possible to create a 20% HCl solution by mixing the given solutions.

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1. The value of  P(A or B) is approximately 0.85.

2. Event A and Event B are not mutually exclusive.

1. To find the probability of the union of two events, A or B, we can use the formula:

P(A or B) = P(A) + P(B) - P(A and B)

Given that P(A) = 0.59, P(B) = 0.80, and P(A and B) = 0.54, we can substitute these values into the formula:

P(A or B) = 0.59 + 0.80 - 0.54

P(A or B) = 0.85

Therefore, P(A or B) is approximately 0.85.

2. To determine if Event A and Event B are mutually exclusive, we need to check if they can both occur at the same time. If the intersection of A and B (P(A and B)) is zero, then they are mutually exclusive.

However, in this case, P(A and B) is not zero (it is 0.54). Therefore, Event A and Event B are not mutually exclusive.

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To find the minimum value of the average cost over the given intervals, we need to calculate the average cost function and evaluate it at the endpoints of each interval.

a) For the interval 1 ≤ x ≤ 10, the average cost function is given by C_avg = (C(10) - C(1))/(10 - 1), where C(x) = x + 30x + 128. Evaluating C(10) and C(1), we get C(10) = 10 + 30(10) + 128 = 388 and C(1) = 1 + 30(1) + 128 = 159. Plugging these values into the average cost function, we have C_avg = (388 - 159)/(10 - 1) = 229/9 ≈ 25.4. Therefore, the minimum value of the average cost over the interval 1 ≤ x ≤ 10 is approximately 25.4.

b) Similarly, for the interval 10 ≤ x ≤ 20, we calculate the average cost function C_avg = (C(20) - C(10))/(20 - 10). Evaluating C(20) and C(10), we get C(20) = 20 + 30(20) + 128 = 748 and C(10) = 10 + 30(10) + 128 = 388. Plugging these values into the average cost function, we have C_avg = (748 - 388)/(20 - 10) = 36. Therefore, the minimum value of the average cost over the interval 10 ≤ x ≤ 20 is 36.

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Discounted price for room would be 70% of the original list price, which is 0.7 * RM450 = RM315. The final amount that Ms. Jessi needs to pay in cash is RM283.50 - RM14.18 = RM269.32, which rounds up to RM269.33.

Ms. Jessi will spend RM256.75 in cash for a room at a list price of RM450 if she qualifies for the trade discount. The hotel reduces the prices of all room types by 30% during the low season. This means the discounted price for the room would be 70% of the original list price, which is 0.7 * RM450 = RM315.

Additionally, Ms. Jessi qualifies for a 10% trade discount, which further reduces the price. The trade discount is calculated as 10% of RM315, which is 0.1 * RM315 = RM31.50. Therefore, the price after applying the trade discount is RM315 - RM31.50 = RM283.50.

Finally, Ms. Jessi is eligible for a 5% cash discount, which is calculated as 5% of RM283.50, resulting in a cash discount of 0.05 * RM283.50 = RM14.18. The final amount that Ms. Jessi needs to pay in cash is RM283.50 - RM14.18 = RM269.32, which rounds up to RM269.33. Therefore, the correct answer is d. RM269.33.

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The sum of the exterior angle of the pentagon is 360 degrees.

The polygon above is a pentagon. A pentagon is a polygon with 5 sides.

If the side of a polygon is extended, the angle formed outside the polygon is the exterior angle. The sum of the exterior angles of a polygon is 360°.

Therefore, the sum of the measure of the exterior angles of the pentagon as shown is 360 degrees.

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(a) The null hypothesis (H0): The proportion of residents in the community who recycle is still 60%.

The alternative hypothesis (Ha): The proportion of residents in the community who recycle is less than 60%.

(b) The appropriate test statistic to use in this case is the z-test for proportions.

(c) To find the value of the test statistic, we need to calculate the standard error (SE) and the z-score.

The formula for the standard error of a proportion is:

SE = √[(p * (1 - p)) / n]

where p is the assumed proportion (60%) and n is the sample size (250). Substituting the values, we get:

SE = √[(0.60 * 0.40) / 250] ≈ 0.0308

Next, we calculate the z-score using the formula:

z = (x - p) / SE

where x is the number of residents in the sample who recycle (129). Substituting the values, we have:

z = (129 - (0.60 * 250)) / 0.0308 ≈ -7.767

(d) The p-value is the probability of observing a test statistic as extreme as the one calculated under the null hypothesis.

Since this is a one-tailed test (looking for evidence of a decrease in the proportion), we need to find the area to the left of the calculated z-score. Consulting a standard normal distribution table or using statistical software, we find that the p-value is essentially 0.

(e) Since the p-value is less than the significance level of 0.05, we reject the null hypothesis. There is enough evidence to support the policy maker's claim that the proportion of residents who recycle is less than 60%.

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A 99% confidence interval for the mean tuition for all colleges and universities in the United States is ($13,885-$22,515). A simple random sample of 40 colleges and universities in the United States has a mean tuition of $18,200 with a standard deviation of $10,600.

To construct a 99% confidence interval for the mean tuition for all colleges and universities in the United States, the steps involved are;

Step 1: Identify the level of confidence and the sample size of the problemLevel of confidence= 99%This indicates that we have a 99% confidence level. Sample size = 40

Step 2: Look up the z-values of a standard normal distribution for the given level of confidence.For a 99% confidence interval, the z-value would be 2.576.

Step 3: Calculate the Standard errorStandard error, SE = σ/ √n, where σ is the standard deviation and n is the sample size.SE= 10600/√40= 1677.5

Step 4: Determine the margin of errorMargin of error = z*SEMargin of error = 2.576 x 1677.5= 4315.14

Step 5: Determine the confidence interval.The confidence interval can be calculated by taking the sample mean and adding and subtracting the margin of error from it.

Confidence interval= $18,200±$4315.14=$13,884.86-$22,515.14

Therefore, a 99% confidence interval for the mean tuition for all colleges and universities in the United States is ($13,885-$22,515).

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To find the number of passengers that will maximize the revenue received from the flight, we need to determine the point at which the revenue is maximized. This can be done by analyzing the cost and refund structure of the trip.

Let's denote the number of passengers as 'n'. For the first 85 passengers, the cost per passenger is $569. For each additional passenger, there is a refund of $5. Therefore, the revenue function can be expressed as R(n) = (569 - 5(n-85))n, where R(n) represents the revenue obtained from 'n' passengers.

To find the number of passengers that maximize the revenue, we need to find the value of 'n' that maximizes the revenue function R(n). We can accomplish this by taking the derivative of R(n) with respect to 'n', setting it equal to zero, and solving for 'n'.

Differentiating R(n) with respect to 'n' gives us dR/dn = 569 - 10(n-85). Setting this derivative equal to zero and solving for 'n' yields 569 - 10(n-85) = 0. Solving this equation, we find n = 79.5.

Since the number of passengers must be a whole number, we round 79.5 to the nearest integer, which is 80. Therefore, the number of passengers that will maximize the revenue received from the flight is 80.

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To find dy/dx, we differentiate both sides of the given equation with respect to x using the rules of differentiation. Applying the chain rule and the power rule, we have: -231x^6 + 297x^32y + 2yy' = 0

Next, we can solve this equation for dy/dx by isolating the derivative term. Rearranging the equation, we get:

dy/dx = (231x^6 - 2yy') / (297x^32)

Now, to find the equation of the tangent line at the point (1, 1), we substitute the coordinates (x, y) = (1, 1) into the derivative expression dy/dx.

Substituting x = 1 and y = 1 into the equation, we get:

dy/dx = (231(1)^6 - 2(1)(y')) / (297(1)^32)

      = (231 - 2y') / 297

Since the point (1, 1) lies on the tangent line, we can substitute x = 1 and y = 1 into the original equation to find y'. We have:

-33(1)^7 + 9(1)^33(1) + (1)^2 = -23

-33 + 9 + 1 = -23

-23 = -23

Thus, y' at (1, 1) is indeterminate. Therefore, we cannot determine the equation of the tangent line in the form y = mx + b without knowing the value of y'

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The seepage rate of this hose is 4.569 L/hour.

The sweat hose has porous walls through which water seeps. The seepage rate of this hose in liters/hour is to be calculated.

The data given for the calculation is as follows:

The hose is 15 m long, 3 cm o.d., and 2 cm i.d.

It is connected to a water faucet at one end, and it is sealed at the other end. It has 100 pores/cm2 based on the outside of the hose surface. Each pore is tubular, 0.5 cm long and 10μm in diameter. The water pressure at the faucet feeding the hose is 100kPa above atmospheric pressure. What is the formula for seepage rate?

The formula for seepage rate is given as,Q = kA(2gh/L)^(1/2)Here,Q = seepage ratek = coefficient of permeabilityA = total area of the soil massg = acceleration due to gravityh = head of water above the soil massL = length of soil massThe required seepage rate can be calculated as follows: Given,Length of the hose, L = 15 mOuter diameter of the hose, d = 3 cmInner diameter of the hose, d_i = 2 cm. Radius of the hose, R = d/2 = 1.5 cm. Radius of the inner surface of the hose, R_i = d_i/2 = 1 cmArea of the outer surface of the hose, A_o = πR^2 = 22.5π cm^2Area of the inner surface of the hose, A_i = πR_i^2 = π cm^2Total area of the soil mass, A = A_o - A_i = 21.5π cm^2Pressure head of water, h = 100 kPaPore diameter, d_p = 10 μm = 0.001 cmPore length, l_p = 0.5 cm = 0.005 mNumber of pores per unit area, n = 100/cm^2 = 10^4/m^2Coefficient of permeability, k = (d_p^2/32)*n*l_p = (0.001^2/32)*10^4*0.005 = 0.001953125 m/sSeepage rate, Q = kA(2gh/L)^(1/2)Q = (0.001953125)*(21.5π)*(2*9.81*100/1000/15)^(1/2) = 4.569 L/hour.

Therefore, the seepage rate of this hose is 4.569 L/hour

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a) The correlation coefficient r ≈ -0.723, letting row 1 represent the x-values and row 2 the y-values.

b) The correlation coefficient r ≈ -1.334, letting row 2 represent the x-values and row 1 the y-values.

c) Switching the variables changes the sign of the correlation coefficient from negative to positive and increases its absolute value.

To calculate the correlation coefficient, you can use the following steps:

Step 1: Find the means (averages) of both x and y values.

x = (16 + 30 + 38 + 45 + 53 + 62 + 80) / 7 = 45.71

y = (144 + 131 + 131 + 201 + 162 + 190 + 134) / 7 = 159.57

Step 2: Subtract the mean of x from each x value and the mean of y from each y value.

xᵢ - x: -29.71, -15.71, -7.71, -0.71, 7.29, 16.29, 34.29

yᵢ - y: -15.57, -28.57, -28.57, 41.43, 2.43, 30.43, -25.57

Step 3: Square each of the differences obtained in Step 2.

(-29.71)², (-15.71)², (-7.71)², (-0.71)², (7.29)², (16.29)², (34.29)²

(-15.57)², (-28.57)², (-28.57)², (41.43)², (2.43)², (30.43)², (-25.57)²

Step 4: Find the sum of the squared differences.

Σ(xᵢ - x)² = 4327.43

Σ(yᵢ - y)² = 17811.43

Step 5: Multiply the corresponding differences from Step 2 for each pair of values and find their sum.

(-29.71)(-15.57), (-15.71)(-28.57), (-7.71)(-28.57), (-0.71)(41.43), (7.29)(2.43), (16.29)(30.43), (34.29)(-25.57)

Σ(xᵢ - x)(yᵢ - y) = -6356.86

Step 6: Calculate the correlation coefficient using the formula:

r = Σ(xᵢ - x)(yᵢ - y) / √[Σ(xᵢ - x)² × Σ(yᵢ - y)²]

r = -6356.86 / √(4327.43 × 17811.43)

r = -6356.86 / √(77117647.5204)

r ≈ -6356.86 / 8777.767

r ≈ -0.723 (rounded to three decimal places)

Now, let's calculate the correlation coefficient when row 2 represents the x-values and row 1 represents the y-values.

Step 1: Find the means (averages) of both x and y values.

x = (144 + 131 + 131 + 201 + 162 + 190 + 134) / 7 = 158.43

y = (16 + 30 + 38 + 45 + 53 + 62 + 80) / 7 = 46.71

Step 2: Subtract the mean of x from each x value and the mean of y from each y value.

xᵢ - x: -14.43, -27.43, -27.43, 42.57, 3.57, 31.57, -24.43

yᵢ - y: -30.71, -16.71, -8.71, -1.71, 6.29, 15.29, 33.29

Step 3: Square each of the differences obtained in Step 2.

(-14.43)², (-27.43)², (-27.43)², (42.57)², (3.57)², (31.57)², (-24.43)²

(-30.71)², (-16.71)², (-8.71)², (-1.71)², (6.29)², (15.29)², (33.29)²

Step 4: Find the sum of the squared differences.

Σ(xᵢ - x)² = 4230.43

Σ(yᵢ - y)² = 3574.79

Step 5: Multiply the corresponding differences from Step 2 for each pair of values and find their sum.

(-14.43)(-30.71), (-27.43)(-16.71), (-27.43)(-8.71), (42.57)(-1.71), (3.57)(6.29), (31.57)(15.29), (-24.43)(33.29)

Σ(xᵢ - x)(yᵢ - y) = -5180.43

Step 6: Calculate the correlation coefficient using the formula:

r = Σ(xᵢ - x)(yᵢ - y) / √[Σ(xᵢ - x)² × Σ(yᵢ - y)²]

r = -5180.43 / √(4230.43 × 3574.79)

r = -5180.43 / √(15111341.6041)

r ≈ -5180.43 / 3887.787

r ≈ -1.334 (rounded to three decimal places)

Switching the variables (x and y) changes the correlation coefficient. In the first calculation, the correlation coefficient (r) is approximately -0.723, and in the second calculation, when the variables are switched, the correlation coefficient (r) is approximately -1.334.

Therefore, switching the variables changes the sign of the correlation coefficient from negative to positive and increases its absolute value.

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To evaluate the given integral using Cauchy's residue theorem, we need to identify the singularities within the contour and calculate their residues. The integral is ∫(z sin z) dz.

To evaluate the integral using Cauchy's residue theorem, we need to identify the singularities of the integrand within the given contours and compute their residues.

(a) For |z| = 3, the singularities are at z = 1 and z = -2 (with multiplicity 2). We calculate the residues at these points and use the residue theorem to evaluate the integral.

(b) For |z| = 3, we need more information about the singularities or the contour to determine the residues and evaluate the integral.

(c) For |z| = 3, the singularities are at z = 0 and z = ∞. We calculate the residues at these points and apply the residue theorem to find the integral value.

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The estimated number of internet hosts in 2021 is approximately 30,735 (rounded to the nearest whole number

To find an exponential regression model for the number of internet hosts from 1994 to 2012, we can use the given data. Using the formula y = ab^x, where x represents the number of years since 1994, we can find the values of a and b that best fit the data. Once we have the regression equation, we can use it to estimate the number of internet hosts in 2021.

To find the exponential regression model for the number of internet hosts, we need to fit the given data to the equation y = ab^x. We can use the data points provided in the table to find the values of a and b.

Using the point (0, 2.6) for the year 1994, we have the equation 2.6 = ab^0, which simplifies to 2.6 = a.

Now, we can use another data point, such as (3, 16.2) for the year 1997, to find the value of b. Substituting the values into the equation, we get 16.2 = 2.6 * b^3. Solving for b, we find b ≈ 1.659.

Therefore, the exponential regression model for the number of internet hosts is given by y = 2.6 * (1.659)^x.

To estimate the number of internet hosts in 2021 (which is 27 years after 1994), we substitute x = 27 into the regression equation:

y = 2.6 * (1.659)^27 ≈ 30734.9566 (rounded to four decimal places).

Thus, the estimated number of internet hosts in 2021 is approximately 30,735 (rounded to the nearest whole number).


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Optimal buffer size for production line can be determined by minimizing sum of capital cost and inventory cost.Without specific values it is not possible to get answers.

The inventory cost is considered as the dollar value of the average buffer level. To calculate the optimal buffer size, we need to compare the costs associated with different configurations of the production line and select the one with the lowest total cost. If the inventory cost is $50 each, we can calculate the total cost for each configuration and select the one with the minimum cost. The cost of the production line includes the cost of the machines and the inventory cost. By comparing the costs for running the line at 1 part per minute and 2 parts per minute, we can determine the optimal buffer size.

To calculate the cost of the optimal line when inventory costs $70 each and $400 each, we follow the same procedure as mentioned above. We compare the costs for different line configurations and select the one with the minimum total cost.

The calculations involve considering the costs of different machine models, line speeds, and inventory costs, and determining the optimal combination that minimizes the total cost. Without specific values for the costs and inventory levels, it is not possible to provide the exact answers. The analysis requires evaluating the costs and selecting the configuration that results in the lowest total cost.

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The particular solution can be found using the method of variation of parameters.

Given that Y is a fundamental matrix for the complementary system, we can use the variation of parameters method to find a particular solution for the given differential equation.

The differential equation is:

Y' + Y = 1 - 2t et

To find a particular solution, we assume the particular solution has the form:

Yp = u₁(t)y₁ + u₂(t)y₂

where y₁ and y₂ are the columns of the fundamental matrix Y, and u₁(t) and u₂(t) are unknown functions to be determined.

We can write the particular solution as:

Yp = u₁(t)[1] + u₂(t)[e²t]

Taking the derivatives, we have:

Yp' = u₁'(t)[1] + u₂'(t)[e²t] + u₂(t)[2e²t]

Substituting these expressions into the differential equation, we get:

u₁'(t)[1] + u₂'(t)[e²t] + u₂(t)[2e²t] + u₁(t)[1] + u₂(t)[e²t] = 1 - 2t et

By comparing the coefficients of the basis functions, we obtain the following system of equations:

u₁'(t) + u₁(t) = 1 - 2t

u₂'(t) + 2u₂(t) = 0

Solving these equations, we can determine the functions u₁(t) and u₂(t). Once we have the functions u₁(t) and u₂(t), we can substitute them back into the particular solution expression to obtain the final particular solution Yp.

Note: The specific solution depends on the values and initial conditions given in the problem, which are not provided.

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The area of the region bounded by y = e^x, y = 12 - e^x, and the y-axis is approximately 23.091 square units.

To find the area of the region, we need to determine the intersection points of the two curves, y = e^x and y = 12 - e^x. By setting the equations equal to each other, we have:

e^x = 12 - e^x

2e^x = 12

e^x = 6

Taking the natural logarithm of both sides, we get:

x = ln(6)

This intersection point serves as the right boundary of the region. The left boundary is the y-axis, which corresponds to x = 0.

To find the area, we integrate the difference of the two curves over the interval [0, ln(6)]. Thus, the area can be calculated as:

A = ∫[0, ln(6)] (12 - e^x - e^x) dx

Simplifying the integrand, we have:

A = ∫[0, ln(6)] (12 - 2e^x) dx

Evaluating the integral, we get:

A = [12x - 2e^x] [0, ln(6)]

A = 12ln(6) - 2(6 - 1)

A ≈ 23.091 square units

Therefore, the area of the region bounded by y = e^x, y = 12 - e^x, and the y-axis is approximately 23.091 square units.

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1) The equation of the regression line is:

y = 0.746x + 55.287

2) The correlation coefficient measures the strength and direction of the linear relationship between the two variables.

3) In this case, about 64% of the variation in final exam scores can be explained by the linear relationship with ca(x) scores.

4) we can reject the null hypothesis and conclude that there is a significant linear relationship between ca(x) and final exam scores in this sample.

5) The predicted final exam score for a student with a CA score of 72 is approximately 103.032 out of 100.

For the equation of the regression, we first need to calculate the means of X and Y, which are 81.44 and 88.6, respectively.

Then we can use the formula for the slope and intercept of the regression line:

b = Σ((xi - x)(yi - y))/Σ(xi - x)²

a = y - bx

where b is the slope, a is the intercept, x is the predictor variable (ca(x)), y is the response variable (final exam scores), xi and yi are the individual values of X and Y, respectively, and Σ is the sum over all values of i.

After performing the calculations, we get:

b = 0.746

a = 55.287

Therefore, the equation of the regression line is:

y = 0.746x + 55.287

To find the correlation coefficient and coefficient of determination, we can use the following formulas:

r = Σ((xi - x)(yi - y))/√(Σ(xi - x)² Σ(yi - y)²)

r² = coefficient of determination

After performing the calculations, we get:

r = 0.799

r² = 0.639

The correlation coefficient measures the strength and direction of the linear relationship between the two variables.

In this case, we have a moderately strong positive correlation, which means that higher ca(x) scores tend to be associated with higher final exam scores.

The coefficient of determination represents the proportion of the variance in Y that can be explained by the regression model.

In this case, about 64% of the variation in final exam scores can be explained by the linear relationship with ca(x) scores.

To test the significance of the model at a = 5%, we can perform a hypothesis test on the slope of the regression line.

The null hypothesis is that the slope is equal to zero (i.e., there is no linear relationship between ca(x) and final exam scores), and the alternative hypothesis is that the slope is different from zero.

We can use a t-test with n-2 degrees of freedom, where n is the sample size (10 in this case).

After performing the calculations, we get a t-value of 3.213, which is greater than the critical value of 2.306 (since we have a two-tailed test and a = 5% with 8 degrees of freedom).

Therefore, we can reject the null hypothesis and conclude that there is a significant linear relationship between ca(x) and final exam scores in this sample.

To calculate the predicted final exam score of a student whose CA is 72, we can use the equation of the regression line:

y = 0.746x + 55.287

where x is the CA score and y is the predicted final exam score.

Substituting x = 72 into the equation, we get:

y = 0.746(72) + 55.287

y = 103.032

Therefore, the predicted final exam score for a student with a CA score of 72 is approximately 103.032 out of 100.

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The value of fy(2, -1) is -2e.a partial derivative of a function of several variables is its derivative with respect to one of those variables, with the others held constant (as opposed to the total derivative, in which all variables are allowed to vary).

Partial derivatives are used in vector calculus and differential geometry.

The partial derivative of a function f(x, y) with respect to x is denoted by ∂f/∂x. The partial derivative of f(x, y) with respect to y is denoted by ∂f/∂y.

The partial derivative of f(x, y) with respect to y is equal to e^x / y^2. To find the value of fy(2, -1), we need to evaluate this partial derivative at the point (2, -1). ∂f/∂y = e¹/y

When x = 2 and y = -1, the value of the partial derivative is equal to -2e. This is because e¹/(-1) = -e.

Therefore, the value of fy(2, -1) is -2e.

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When using the common summations, such as the first n squared natural numbers, the lower limit of summation must be 0.

The answer to the question "When using the common summations, such as first n squared natural numbers, the lower limit of summation must be" is option a, 0. This is because when using summation notation, the lower limit represents the first term in the series. For the first n squared natural numbers, the series would start at 1, so the lower limit would be 1. However, in many cases, the series starts at 0 and goes up to n-1. For example, the summation of the first n natural numbers would be written as: ∑ i=0^n-1 i. Here, the series starts at 0 and goes up to n-1. Summation notation is a mathematical shorthand that allows us to express large series of numbers more compactly. It is especially useful for expressing infinite series, which would otherwise be impossible to write out fully. The notation involves the use of a sigma symbol (Σ) to indicate a series, followed by an expression that describes the terms of the series. This expression is written to the right of the sigma symbol and includes an index variable, which tells us which term we are currently evaluating. For example, the sum of the first n natural numbers can be written as: ∑ i=1^n i. Here, the index variable is i, and it ranges from 1 to n, indicating that we are adding up all the natural numbers from 1 to n.One use of summation notation is to simplify mathematical expressions and write them more compactly. By using this notation, we can express large series of numbers in a concise and elegant way, making it easier to work with them. We can also use summation notation to express more complicated mathematical concepts, such as geometric series, trigonometric series, and so on. This notation is especially useful in calculus, where we often encounter infinite series that are difficult to evaluate by hand. With summation notation, we can express these series more clearly and see how they behave as we approach infinity.

The answer to the first question is a) 0 and summation notation is a shorthand for writing series of numbers in a compact way. It is used to simplify mathematical expressions and make it easier to work with large series of numbers. Summation notation is especially useful for expressing infinite series, which would otherwise be impossible to write out fully.

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1. The standard error of the mean for samples of size 20 is approximately 7.16. 2. The 95% confidence interval is: (lower limit, upper limit) = (66 - 13.94, 66 + 13.94) ≈ (52.06, 79.94) (rounded to 2 decimal places). 3. The lower limit of the 95% confidence interval is approximately 52.06. 4. The width of the 95% confidence interval is approximately 27.88.

1. The standard error of the mean for samples of size 20 can be calculated using the formula:

Standard Error = s / sqrt(n)

where s is the known standard deviation of the population and n is the sample size.

In this case, s = 32 and n = 20. Substituting the values into the formula, we have:

Standard Error = 32 / sqrt(20) ≈ 7.16 (rounded to 3 decimal places)

Therefore, the standard error of the mean for samples of size 20 is approximately 7.16.

2. The 95% confidence interval for the population mean can be calculated using the formula:

(lower limit, upper limit) = (sample mean - margin of error, sample mean + margin of error)

The margin of error is determined by the critical value of the t-distribution at a 95% confidence level and the standard error of the mean.

Since the sample size is 20, the degrees of freedom for the t-distribution will be 20 - 1 = 19.

Using a t-table or calculator, the critical value for a 95% confidence level with 19 degrees of freedom is approximately 2.093.

The margin of error is calculated as:

Margin of Error = critical value * standard error = 2.093 * (s / sqrt(n)) = 2.093 * (32 / sqrt(20)) ≈ 13.94 (rounded to 2 decimal places)

Therefore, the 95% confidence interval is:

(lower limit, upper limit) = (66 - 13.94, 66 + 13.94) ≈ (52.06, 79.94) (rounded to 2 decimal places)

3. The lower limit of the 95% confidence interval is approximately 52.06.

4. The width of the confidence interval can be calculated by subtracting the lower limit from the upper limit:

Width of Confidence Interval = upper limit - lower limit = 79.94 - 52.06 ≈ 27.88 (rounded to 2 decimal places)

Therefore, the width of the 95% confidence interval is approximately 27.88.

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If 28% of adult workers have a high school diploma but do not pursue any further education, and 365 adult workers are randomly selected, we would expect 102.2 adult workers to have a high school diploma but do not pursue any further education.

This is calculated using the following formula:

Expected value = n * p

where:

n is the number of trials

p is the probability of success

In this case, n = 365 and p = 0.28. Therefore, the expected value is:

Expected value = 365 * 0.28 = 102.2

It is important to note that this is just an expected value. The actual number of adult workers who have a high school diploma but do not pursue any further education may be more or less than 102.2.

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The given series ∑ n=1 [infinity] (43π)^n can be determined to converge or diverge using appropriate tests. The p⋅ Series Test and the Geometric Series Test can be applied to analyze the convergence behavior.

The series ∑ n=1 [infinity] (43π)^n is a geometric series with a common ratio of 43π. The Geometric Series Test states that a geometric series converges if the absolute value of the common ratio is less than 1 and diverges otherwise.

In this case, since the absolute value of the common ratio 43π is greater than 1, the series diverges by the Geometric Series Test.

Therefore, the correct answer is that the given series ∑ n=1 [infinity] (43π)^n diverges by the Geometric Series Test.

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The function has a local minimum at x = -56/17 and local maximums at x = -7 and x = 0.

Given the functions, we are supposed to find the interval(s) at which the functions are increasing, decreasing, local minimum, and local maximum. The functions are given below:

1) f(x) = -4x³ - 6.72x² + 379.3068x + 2.44

To find the interval(s) where the function is increasing or decreasing, we can differentiate the given function and find the critical point(s). Then, we can use the first derivative test to determine the intervals where the function is increasing or decreasing. We can then use the second derivative test to find the interval(s) where the function has local minimum and local maximum.

Now, let's differentiate the given function to get its first and second derivatives.

f(x) = -4x³ - 6.72x² + 379.3068x + 2.44

Differentiating with respect to x, we get f'(x) = -12x² - 13.44x + 379.3068

Now, we need to find the critical point(s). To do so, we will equate the first derivative to zero and solve for x.

f'(x) = 0 => -12x² - 13.44x + 379.3068 = 0

Solving the above equation using the quadratic formula, we get

x = (-b ± √(b² - 4ac))/(2a) = (-(-13.44) ± √((-13.44)² - 4(-12)(379.3068)))/(2(-12)) = (13.44 ± √(13.44² + 4*12*379.3068))/(2*12)

= (13.44 ± √18905.8769)/24 ≈ 12.611 or -10.132

Therefore, the critical points are x = 12.611 and x = -10.132.

Now, we can use the first derivative test to find the intervals where the function is increasing or decreasing. We will consider the intervals separated by the critical points.

Therefore, the given function is increasing on the interval (-10.132, 12.611) and decreasing on the intervals (−∞, −10.132) and (12.611, ∞).

Now, we can find the local minimum and maximum of the function on these intervals using the second derivative test. For this, we need to find the second derivative of the function. Differentiating the first derivative with respect to x, we get f''(x) = -24x - 13.44

The second derivative is negative for x < -10.132, positive for -10.132 < x < 12.611, and negative for x > 12.611.

Therefore, the function has a local maximum at x = -10.132 and a local minimum at

x = 12.611.2) f(x) = x³(x + 7)⁸, for x greater than or equal to -13 and less than or equal to 15.

To find the interval(s) where the function is increasing or decreasing, we can differentiate the given function and find the critical point(s).

Then, we can use the first derivative test to determine the intervals where the function is increasing or decreasing. We can then use the second derivative test to find the interval(s) where the function has local minimum and local maximum.

Now, let's differentiate the given function to get its first and second derivatives. f(x) = x³(x + 7)⁸

Differentiating with respect to x, we get

f'(x) = 9x²(x + 7)⁷ + x³*8(x + 7)⁶= x²(x + 7)⁶(9x + 8x + 56)

Now, we need to find the critical point(s). To do so, we will equate the first derivative to zero and solve for x.

f'(x) = 0 => x²(x + 7)⁶(9x + 8x + 56) = 0

Therefore, the critical points are x = 0, x = -7, and x = -56/17.

Now, we can use the first derivative test to find the intervals where the function is increasing or decreasing.

Therefore, the given function is increasing on the intervals (-13, -56/17) and (0, 15) and decreasing on the interval (-7, 0).

Now, we can find the local minimum and maximum of the function on these intervals using the second derivative test. For this, we need to find the second derivative of the function.

Differentiating the first derivative with respect to x, we get

f''(x) = 54x(x + 7)⁶ + 18x²(x + 7)⁵ + 2x³(x + 7)⁴

The second derivative is positive for x < -7, negative for -7 < x < -56/17, and positive for x > -56/17.

Therefore, the function has a local minimum at x = -56/17 and local maximums at x = -7 and x = 0.

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We can be 95% confident that the true population mean of the number of hours per week a teacher spends working at home falls within the interval (8.10, 8.90) hours per week.

To construct a confidence interval for the mean number of hours per week a teacher spends working at home, we can use the following steps:

Step 1: Identify the necessary information:

- Sample mean ([tex]\bar{X}[/tex]) = 8.5 hours per week

- Sample size (n) = 52

- Population standard deviation (σ) = 1.5 hours per week

- Confidence level = 95%

Step 2: Determine the critical value:

Since the sample size is relatively large (n > 30) and the population standard deviation is known, we can use the Z distribution. At a 95% confidence level, the critical value corresponds to a two-tailed test, with α/2 = 0.025. Looking up the critical value in the Z-table, we find it to be approximately 1.96.

Step 3: Calculate the margin of error:

The margin of error (E) is given by the formula: E = z * (σ / √n), where z is the critical value, σ is the population standard deviation, and n is the sample size. Substituting the values, we have:

E = 1.96 * (1.5 / √52)

Step 4: Calculate the confidence interval:

The confidence interval can be calculated as: Confidence Interval = [tex]\bar{X}[/tex] ± E, where [tex]\bar{X}[/tex] is the sample mean and E is the margin of error.

Confidence Interval = 8.5 ± E

Step 5: Interpret the confidence interval:

The confidence interval represents the range of values within which we can be confident (at a certain confidence level) that the true population mean lies. In this case, the 95% confidence interval for the mean number of hours per week a teacher spends working at home is given by:

Confidence Interval = 8.5 ± E

Now, let's calculate the margin of error (E) and the confidence interval:

E = 1.96 * (1.5 / √52) ≈ 0.4035

Confidence Interval = 8.5 ± 0.4035

Interpretation: We can be 95% confident that the true population mean of the number of hours per week a teacher spends working at home falls within the interval (8.10, 8.90) hours per week.

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Complete question is below

This activity will allow you to explore on finding and interpreting confidence intervals for both a population mean and a population proportion. Read the steps below and complete each item.

Instructor Ramos is concerned about the amount of time teachers spend each week doing schoolwork at home. A simple random sample of 52 teachers had a mean of 8.5 hours per week working at home after school. Construct and interpret a 95% confidence interval for the mean number of hours per week a teacher spends working at home. Assume that the population standard deviation is 1.5 hours per week