Grades on a very large statistics course have historically been awarded according to the following distribution. HD D С P Z or Fail 0.15 0.20 0.30 0.30 0.05 What is the probability that two students picked independent of each other and at random both get a Z?a. 0.0100 b. 0.0225 c. 0.0500 d. 0.0025

When the weekly sales are x = 54 units, we must get the derivative of the profit function with respect to time in order to determine the rate of change in sales with respect to time. Pwhere C is the

The term "rate of change" describes how quickly a variable changes over time. It gauges how much a variable alters over the course of a certain length of time. The derivative of a function in mathematics serves as a symbol for pace of change. A function's derivative shows how quickly a function changes at any given point on its graph. Numerous real-world events, such as changes in temperature, velocity, and stock prices, may be studied using the rate of change. A moving object's acceleration is calculated in physics, while the rate of return on an investment is calculated in finance. A helpful tool for studying change is the rate of change

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complete question:

The profit for a product is increasing at a rate of $5800 per week. The demand and cost functions for the product are given by p = 8000 − 25x and C = 2400x + 54, where x is the number of units produced per week. Find the rate of change of sales with respect to time when the weekly sales are x = 54 units.

_____?_____ units per week

The probability that Brandon ordered a pepperoni pizza is 1/6.

To find the probability that Brandon ordered a pepperoni pizza, we need to first determine the fraction of pizzas sold that were pepperoni.

We know that 1/6 of the pizzas sold were cheese, which means that 5/6 of the pizzas sold were not cheese. So, if we let x be the total number of pizzas sold in the week, then (5/6)x is the number of pizzas sold that were not cheese.

Of those non-cheese pizzas, 1/5 were pepperoni. So the total number of pepperoni pizzas sold would be (1/5)(5/6)x = (1/6)x.

Therefore, the probability that Brandon ordered a pepperoni pizza is (1/6)x / x = 1/6.

So the answer is: The probability that Brandon ordered a pepperoni pizza is 1/6.

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By conducting this study, you will be able to identify any areas for improvement in the measurement system or operator training. This will help to ensure that the measurements are consistent and accurate, ultimately leading to a better quality product.

To assess measurement system variation among operators using handheld calipers to measure wooden floorboards, you conducted a crossed-gage R&R study using MINITAB software. You had 3 operators use the same calipers to randomly measure 10 wooden floorboards twice, resulting in a total of 60 measurements. The data was stored in a MINITAB worksheet called Floor Board.mwx.

The graphical output of the crossed-gage R&R study will show the amount of variation that is due to the measurement system, as well as the amount of variation that is due to the operators themselves. This will allow you to identify any issues with the measurement system or operator training that may be contributing to the measurement variation.

In MINITAB, you can analyze the data using the crossed gage R&R tool. This will calculate the measurement system variation, operator variation, and the total variation. The results can be presented in a graph or table format, allowing you to easily compare the different sources of variation.

By conducting this study, you will be able to identify any areas for improvement in the measurement system or operator training. This will help to ensure that the measurements are consistent and accurate, ultimately leading to a better quality product.

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Answer:

-3/2

Step-by-step explanation:

slope = Δy/Δx = (-1 - 2) / (1 - -1) = -3/2 = -1.5

After baking the cookies there will be fractional number 1 3/4 cups of flour will be left.

What is fraction?

Fraction is a part of any whole number. If an object or any thing will be divided into some parts then the parts will be the fraction of the whole thing. There are two parts in a fraction one is numerator another is denominator. Some examples of fractions are 5/2, 7/9 etc.

The quotient Larissa has 4 1/2 cups of flour. She is making cookies using a recipe that calls for 2 3/4 cups of flour.

So the total amount of flour is 4 1/2 cups = 9/2 cups which is a fraction.

The recipe calls for 2 3/4 cups of flour= 11/4 cups which is also a fraction.

Subtracting two fractional terms we will get the result.

            9/2- 11/4

The least common multiple between 9/2 and 11/4 is 4

So using the subtraction property of  fraction  we get   [tex]\frac{18-11}{4}[/tex]  = 7/4

The fraction 7/4 is equivalent to 1 3/4.

Hence , after baking the cookies 1 3/4 cups of flour will be left.

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The value of function are,

⇒ f' (x) = ( sin x - x cos x) - 5 sec² x

And, ⇒ y' = 45x² - 20x + 3

Given that;

Function is,

⇒ f(x) = x sin x - 5 tan x

And, y = (3x - 2) (5x² + 1)

Now, We can simplify as;

⇒ f(x) = x sin x - 5 tan x

Differentiate as;

⇒ f' (x) = ( sin x - x cos x) - 5 sec² x

And, For y = (3x - 2) (5x² + 1)

Differentiate as;

⇒ y' = (3x - 2) (10x) + (5x² + 1) (3)

⇒ y' = 30x² - 20x + 15x² + 3

⇒ y' = 45x² - 20x + 3

Thus, The value of function are,

⇒ f' (x) = ( sin x - x cos x) - 5 sec² x

And, ⇒ y' = 45x² - 20x + 3

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The result is sometimes statistically significant. The correct option is b. sometimes.

Statistical significance is determined by comparing the observed value (in this case, FDATA) to a predetermined threshold, typically referred to as the alpha level or significance level. If the observed value exceeds the alpha level, then the result is considered statistically significant, meaning that the observed value is unlikely to have occurred by chance alone.

In this case, the given value of FDATA is 0.9. However, without knowing the context of the statistical analysis being conducted, it is not possible to determine whether this value is statistically significant or not. The determination of statistical significance depends on various factors, such as the sample size, the research question, the type of statistical test being used, and the desired level of confidence.

Therefore, without additional information about the specific context and analysis being performed, it is not possible to definitively state whether a value of FDATA = 0.9 is statistically significant or not. The result could be statistically significant in some situations (when compared to an appropriate alpha level), and not statistically significant in other situations.

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a. The sample size for this test should be at least 24.

b. Sample deviation rate = 0.1667 or 16.67%

c. The upper deviation limit for the test is 38.6%.

d. A conclusion on whether the control is operating effectively

or not, we compare the sample deviation rate to the tolerable deviation

rate and the upper deviation limit.

a. To determine the sample size for the test, we can use the formula:

[tex]n = (Z^2 \times p \times (1-p)) / d^2[/tex]

where:

Z = the Z-value for the desired level of confidence, which is typically 1.65 for a 90% confidence level

p = the expected deviation rate

d = the tolerable deviation rate -the maximum acceptable deviation rate

Plugging in the values given, we get:

[tex]n = (1.65^2 \times 0.02 \times 0.98) / 0.06^2[/tex]

n = 23.76

b. The sample deviation rate can be calculated by dividing the number of deviations found in the sample by the sample size:

Sample deviation rate = Number of deviations / Sample size

Sample deviation rate = 4 / 24

Sample deviation rate = 0.1667 or 16.67%

c. The upper deviation limit can be calculated using the formula:

UDL = Sample deviation rate + (Z × √((Sample deviation rate × (1 - Sample deviation rate)) / Sample size))

where:

Z = the Z-value for the desired level of confidence, which is 1.65 for a 90% confidence level

Plugging in the values given, we get:

UDL = 0.1667 + (1.65 × √((0.1667 × (1 - 0.1667)) / 24))

UDL = 0.386

d. To draw a conclusion on whether the control is operating effectively

or not, we compare the sample deviation rate to the tolerable deviation

rate and the upper deviation limit.

In this case, the sample deviation rate (16.67%) is below the tolerable

deviation rate (6%) and also below the upper deviation limit (38.6%). This

suggests that the control is operating effectively and there is no

significant risk of incorrect acceptance.

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So the equation for the unit rate, r, is:

r = 50/9

Calculating the sum of two or more numbers is the process of multiplication. 'A' multiplied by 'B' is how you express the multiplication of two numbers, let's say 'a' and 'b'. Multiplication in mathematics is essentially just adding a number repeatedly in relation to another number.

To find the unit rate, we need to determine how much of a wall Steven can paint in one hour. We can start by using the information given to find out how much of a wall he can paint in 1/10 of an hour:

1/6 of a wall in 3/10 of an hour

= (1/6) ÷ (3/10)

= (1/6) × (10/3)

= 10/18

= 5/9

Therefore, Steven can paint 5/9 of a wall in 1/10 of an hour.

To find out how much of a wall he can paint in one hour, we can multiply this by 10:

(5/9) × 10 = 50/9

Therefore, Steven can paint 50/9 of a wall in one hour.

So the equation for the unit rate, r, is:

r = 50/9

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(a) A normal quantile plot of the data is illustrated below.

(b) There are some slight deviations from the line, particularly for the data points at the extremes of the distribution (68 and 82 inches).

In this case, we have the heights of 7 NBA players: 68, 69, 78, 82, 75, 73, and 80 inches. To construct the normal quantile plot, we first calculate the mean and standard deviation of the data:

Mean = (68 + 69 + 78 + 82 + 75 + 73 + 80) / 7 = 75.43

Standard deviation = 5.98

Next, we calculate the z-scores for each data point:

z1 = (68 - 75.43) / 5.98 = -1.24

z2 = (69 - 75.43) / 5.98 = -1.07

z3 = (78 - 75.43) / 5.98 = 0.43

z4 = (82 - 75.43) / 5.98 = 1.09

z5 = (75 - 75.43) / 5.98 = -0.07

z6 = (73 - 75.43) / 5.98 = -0.41

z7 = (80 - 75.43) / 5.98 = 0.75

We can then plot the z-scores against the corresponding quantiles of a standard normal distribution. The z-score of -1.24 corresponds to the 10th percentile of the standard normal distribution, while the z-score of 1.09 corresponds to the 86th percentile.

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The solution of the differentiation equation is Yₓ = (-1/3)t² - (1/2)t - 9/40eˣ

In this case, we will guess that the particular solution takes the form of Yₓ = At² + Bt + C, where A, B, and C are constants that we need to find.

To find these constants, we will need to differentiate the solution Yₓ twice and plug it into the differential equation. First, let's find the first derivative of Yₓ:

Yₓ' = 2At + B

Next, let's find the second derivative of Yₓ:

Yₓ'' = 2A

Now, we can plug Yₓ, Yₓ', and Yₓ'' into the differential equation:

13.5e⁻ˣ(2A) + 2(At² + Bt + C) + (At² + Bt + C) = t² + 1

Simplifying this equation gives:

(13.5e⁻ˣ)(2A) + (2A + 1)At² + (2B + 1)Bt + 2C = t² + 1

Now, we can equate the coefficients of each term on both sides of the equation to find the values of A, B, and C.

Starting with the coefficient of t² on both sides, we get:

(13.5e⁻ˣ)(2A) + (2A + 1)A = 1

Simplifying this equation gives:

A = -1/3

Next, we can look at the coefficient of t on both sides:

(2B + 1)B = 0

This equation tells us that either B = 0 or B = -1/2. However, if we set B = 0, then the coefficient of t² on the left side of the equation will be 0, which is not equal to the coefficient of t² on the right side of the equation. Therefore, we must choose B = -1/2.

Finally, we can look at the constant term on both sides:

(13.5e⁻ˣ)(2A) + (2A + 1)C + 2C = 1

Substituting the values of A and B that we found earlier, we get:

(13.5e⁻ˣ)(-2/3) - 1/3C = 0

Simplifying this equation gives:

C = -9/40eˣ

Therefore, our particular solution Yₓ is:

Yₓ = (-1/3)t² - (1/2)t - 9/40eˣ

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The probability of X being less than 70 is approximately 0.0006.

The standard normal distribution.

Transform X into a standard normal variable Z:

[tex]Z = (X - \mu) / \sigma[/tex]

Substituting the given values, we have:

[tex]Z = (70 - 83) / 4 = -3.25[/tex]

Using a standard normal table or calculator, we can find

The probability:

[tex]P(X < 70) = P(Z < -3.25) = 0.0006[/tex]

The probability of X being less than 70 is approximately 0.0006.

the usual distribution of normals.

Change X into the typical normal variable Z:

Z = (X - \mu) / \sigma

If we substitute the values provided, we get:

Z = (70 - 83) / 4 = -3.25

We may determine the probability using a calculator or a normal table to find:

Z = (70 - 83) / 4 = -3.25

X has a about 0.0006 likelihood of being less than 70.

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The expected percent of business majors who dropped in this sample would be 28%.

To find the expected percent of business majors who dropped in this sample, we need to first calculate the total number of students in the sample. From the table below, we see that a total of 250 students dropped general psychology during the fall semester.

| Major              | Number of Students Dropping |
|--------------------|-----------------------------|
| Education          | 20                          |
| Business           | 70                          |
| Arts and Sciences  | 120                         |
| Undecided          | 40                          |
| **Total**             | **250**                        |

Next, we need to calculate the expected number of students who would have dropped from each major if there were no difference among the majors. To do this, we can multiply the total number of students who dropped (250) by the percentage of students in each major:

Education: 0.08 x 250 = 20
Business: 0.28 x 250 = 70
Arts and Sciences: 0.42 x 250 = 105
Undecided: 0.22 x 250 = 55

So, if the university expected that there should be no difference among the different majors in dropping this class, the expected percent of business majors who dropped in this sample would be:

Expected percent of business majors who dropped = (Expected number of business majors who dropped / Total number of students who dropped) x 100
= (70/250) x 100
= 28%

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The answer to this question is 5/2

It will take approximately 8 months to pay off the loan (rounded up to the nearest month).

a. After 6 months, the student will owe $5,383.38 (to the nearest cent).

To calculate this, we can use the formula:

A = P(1 + r/n)^(nt) - PMT[((1 + r/n)^(nt) - 1) / (r/n)]

where:
A = the remaining balance after 6 months
P = the initial loan amount ($8,000)
r = annual percentage rate (3% or 0.03)
n = number of times compounded in a year (12 since it is compounded monthly)
t = time in years (6 months is 0.5 years)
PMT = the monthly payment ($1,000)

Plugging in these values, we get:

A = 8,000(1 + 0.03/12)^(12*0.5) - 1,000[((1 + 0.03/12)^(12*0.5) - 1) / (0.03/12)]

A = $5,383.38 (rounded to the nearest cent)

b. To find out how many months it will take to pay off the loan, we need to keep making the monthly payments until the remaining balance is $0.

Using the same formula as above, we can solve for t:

8,000(1 + 0.03/12)^(12t) - 1,000[((1 + 0.03/12)^(12t) - 1) / (0.03/12)] = 0

Simplifying this equation, we get:

t = log(1 + (1,000/8,000)(0.03/12)) / (12 log(1 + 0.03/12))

t = 7.46 months

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The chance that the average of 49 boxes weighs less than 27.5 ounces is smaller.

We can solve this problem using the central limit theorem. Since the weights of individual boxes have a normal distribution with mean 28 ounces and standard deviation 2 ounces, the distribution of sample means of 49 boxes will also be normal with mean 28 ounces and standard deviation (2/√49) ounces, which simplifies to 0.2857 ounces.

(a) To find the chance that one box has a weight less than 27.5 ounces, we can standardize the weight using the formula z = (x - μ) / σ, where x is the weight we are interested in, μ is the mean weight (28 ounces), and σ is the standard deviation (2 ounces).

So,

z = (27.5 - 28) / 2 = -0.25

Using a standard normal distribution table or calculator, we find that the probability of a z-score less than -0.25 is approximately 0.4013.

(b) To find the chance that 49 boxes have an average weight less than 27.5 ounces, we can standardize the sample mean using the formula z = (x - μ) / (σ / √n), where x is the sample mean weight (27.5 ounces), μ is the mean weight (28 ounces), σ is the standard deviation (2 ounces), and n is the sample size (49 boxes).

So,

z = (27.5 - 28) / (2 / √49) = -1.75

Using a standard normal distribution table or calculator, we find that the probability of a z-score less than -1.75 is approximately 0.0401.

Comparing the two probabilities, we can see that the chance that 49 boxes have an average weight less than 27.5 ounces (0.0401) is smaller than the chance that one box has a weight less than 27.5 ounces (0.4013).

Therefore, the answer is (a) The chance that the average of 49 boxes weighs less than 27.5 ounces is smaller.

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In order to identify the t critical in the t distribution, you’ll need df, # of tails, and alpha

The correct answer is b.

In order to identify the t critical in the t distribution, you'll need the degrees of freedom (df), the number of tails, and alpha. The degrees of freedom are related to the sample size and are necessary to calculate the t statistic. The number of tails refers to whether the test is one-tailed or two-tailed, and alpha is the significance level or probability of rejecting the null hypothesis.

. The one-tailed test is used when the null hypothesis is rejected only when the test results fall on the tails of the distribution. If the test results are in any direction of the distribution, a two-tailed test is used while rejecting the null hypothesis.

Therefore, to determine the critical value in the distribution, we need to know the degrees of freedom (df), the significance level (alpha), and the mantissa (one-tailed or double-tailed). The answer containing these three parameters is b. df, tail count and alpha.

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Check the picture below.

The number of oats that can be used to make 150 cookies is 800 g of oats.

A proportion in mathematics is a claim that two ratios are equal. The relationship between two quantities is called a ratio, and it is typically stated as a fraction.

Geometry, physics, and finance are just a few of the mathematical and science fields where proportions are applied. They are particularly helpful for forming predictions based on patterns or relationships seen, and for comparing two quantities that are in different units or scales.

We can see that;

80 g of oats can be used to produce 15 cookies

x g of oats can produce 150 cookies

x = 80 * 150/15

= 800g

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Answer:

the equation of the line that passes through (3, 14) and is parallel to the line that passes through (10, 2) and (25, 15) is y = (13/15)x + 10.

Step-by-step explanation:

Parallel line equation.

Piyush Soni

Write an equation for the line that passes through (3,14) and is parallel to the line that passes through (10,2) and (25,15)

To find the equation of the line that passes through (3, 14) and is parallel to the line that passes through (10, 2) and (25, 15), we first need to find the slope of the line passing through (10, 2) and (25, 15), which we will call m1.

The slope of the line passing through two points (x1, y1) and (x2, y2) can be calculated using the formula:

m = (y2 - y1) / (x2 - x1)

So, for the line passing through (10, 2) and (25, 15), we have:

m1 = (15 - 2) / (25 - 10) = 13/15

Since we want a line parallel to this one, the slope of our new line will be the same. Let's call this slope m2.

m2 = 13/15

Now, we can use the point-slope form of the equation of a line to find the equation of the line passing through (3, 14) with slope m2:

y - y1 = m2(x - x1)

where x1 = 3 and y1 = 14

Plugging in the values, we get:

y - 14 = (13/15)(x - 3)

Simplifying, we get:

y = (13/15)x + 50/5

or

y = (13/15)x + 10

Therefore, the equation of the line that passes through (3, 14) and is parallel to the line that passes through (10, 2) and (25, 15) is y = (13/15)x + 10.

The equation of the line that passes through (3,14) and is parallel to the line that passes through (10,2) and (25,15) is y = (13/15)x  171/15.

The equation of a line is given by:

y = mx + c

where m is the slope of the line and c is the y-intercept.

Example:

The slope of the line y = 2x + 3 is 2.

The slope of a line that passes through (1, 2) and (2, 3) is 1.

We have,

The slope of a line passing through two points (x1, y1) and (x2, y2) is given by:

slope = (y2 - y1) / (x2 - x1)

Using the points (10,2) and (25,15), we have:

slope = (15 - 2) / (25 - 10) = 13 / 15

Since the line we want is parallel to this line, it will have the same slope. So, we can use the point-slope form of the equation of a line:

y - y1 = m(x - x1)

where m is the slope, and (x1, y1) is the given point (3,14).

Substituting m = 13/15 and (x1,y1) = (3,14).

y - 14 = (13/15)(x - 3)

Expanding and rearranging.

15y - 210 = 13x - 39

15y = 13x - 39 + 210

15y = 13x + 171

y = (13/15)x  171/15

Thus,

The equation of the line that passes through (3,14) and is parallel to the line that passes through (10,2) and (25,15) is y = (13/15)x  171/15.

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Answer:

a. 61%

b. $1315.77

Step-by-step explanation:

a. find percent = part/whole = 732/1200 = 0.61 = 61%

b. 61% of $2157 = 2157 x 0.61 = $1315.77

Answer: 1 2/10

Step-by-step explanation:

The formula you have given (SS₁ + SS₂) / (n₁+ n₂) is actually the formula for the unweighted average of the variances, which is not appropriate when the sample sizes and variances are different between the two samples.

The formula for pooled variance is:

pooled variance = (SS₁+ SS₂) / (df₁ + df₂)

where SS₁ and SS₂ are the sum of squares for the two samples, df₁ and df₂ are the corresponding degrees of freedom, and the pooled variance is the weighted average of the variances of the two samples, where the weights are proportional to their degrees of freedom.

Note that the denominator is df₁ + df₂ not n₁+ n₂. The degrees of freedom take into account the sample sizes as well as the number of parameters estimated in

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If the histogram shows a bell-shaped curve and the normality test (if performed) supports the normality assumption, it appears safe to assume that the normality condition is satisfied for the wood pieces prepared for the lab session, considering them as an SRS of all similar pieces of Douglas fir.

To determine if the normality condition is satisfied, you can follow these steps:

1. Organize the data: Collect the measurements for the characteristics of the wood pieces in your sample (such as density, strength, etc.) and organize them in a list or a table.

2. Create a frequency distribution: Calculate the frequencies of the different measurements and arrange them in a frequency distribution table.

3. Plot a histogram: Using the frequency distribution, create a histogram to visually represent the data. The x-axis represents the measurements and the y-axis represents the frequency.

4. Evaluate the shape of the histogram: Examine the shape of the histogram to determine if it resembles a normal distribution. A normal distribution is characterized by a bell-shaped curve, which is symmetrical around the mean value.

5. Conduct a normality test (optional): If you want to statistically confirm the normality of the data, you can perform a normality test, such as the Shapiro-Wilk test or the Kolmogorov-Smirnov test.

For the wood pieces manufactured for the lab session, using them as an SRS of all comparable pieces of Douglas fir, it is acceptable to infer that the normality criterion is satisfied if the histogram displays a bell-shaped curve and the normality test (if performed) confirms the normality assumption.

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The maximum number of edges in a bipartite graph with two disjoint sets of vertices A (with m elements) and B (with n elements) is m * n.

In the mathematical field of graph theory, a bipartite graph (or bigraph) is a graph whose vertices can be divided into two disjoint and independent sets and , that is every edge connects a vertex in to one in .

To find the maximum number of edges in a bipartite graph with two disjoint sets of vertices A and B, where A has m elements and B has n elements, you can simply multiply the number of elements in set A by the number of elements in set B.

The maximum number of edges in this bipartite graph is given by the product of the sizes of the two vertex sets, which is m * n.

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The current would be 1.5 feet per second fast at a depth of two feet

Drawing the line of best fit, we have the following points

(0, 1.8) and (10, 0.5)

The equation is represented as

y = mx + c

Where

c = y when x = 0

So, we have

y = mx + 1.8

Using the points, we have

10m + 1.8 = 0.5

So, we have

m = -0.13

So, the equation is

y = -0.13x + 1.8

At two feet deep, we have

y = -0.13 * 2 + 1.8

Evaluate

y = 1.54

Approximate

y = 1.5

Hence, the velocity is 1.5 feet per second

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a. X = The length of life of a Sunshine blu-ray player measured in years. Option 4 is the correct answer.

b. The distribution of X is a normal distribution with a mean of 4.1 years and a standard deviation of 1.3 years.

a. The correct definition for the random variable X in this context is 3. X represents the length of time that a Sunshine blu-ray player lasts, measured in years. It is a continuous variable because it can take on any value within a certain range.

b. The distribution of X is a normal distribution, also known as a Gaussian distribution or bell curve. The mean of the distribution is 4.1 years, which is the average length of time that a Sunshine blu-ray player is expected to last. The standard deviation is 1.3 years, which measures the variability or spread of the data. This means that most of blu-ray players will last between approximately 2.8 and 5.4 years, with a smaller number lasting longer or shorter than this range.

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The question is -

A Sunshine blu-ray player is guaranteed for three years. The life of Sunshine blu-ray players is normally distributed with mean of 4.1 years and a standard deviation of 1.3 years. We are interested in the length of time a blu-ray player lasts.

a. Define the random variable X in words.

1. X = The mean length of life of a Sunshine blu-ray player measured in years

2. X = the number of Sunshine blu-ray players that fail in a year

3. X = The length of life of a Sunshine blu-ray player measured in years

4. X = the mean number of Sunshine players sold in a year

b. Describe the distribution of X.

a) The average height of female students in the sample is estimated to be 66.16 inches.

b) The calculated t-value of -4.07 is less than the critical value of -1.66, we reject the null hypothesis in favor of the alternative.

(a) The intercept of 71.0 represents the average height of male students in the sample.

The slope of -4.84 represents the difference in the average height between male and female students. Specifically, the slope implies that, on average, females are 4.84 inches shorter than males.

To estimate the average height of female students, we can set BFemme to 1 in the regression equation:

Female Students: Studenth = 71.0 - 4.84(1) = 66.16 inches.

(b) To test the hypothesis that females, on average, are shorter than males, we can perform a t-test for the coefficient on BFemme.

H0: β1 = 0 (there is no difference in height between males and females)

Ha: β1 < 0 (females are shorter than males)

The t-statistic for the coefficient on BFemme is given by:

[tex]t = (-4.84 - 0) / \sqrt{[(0.3^2 / 29) + (0.57^2 / 81)]} = -4.07[/tex]

where 0.3 and 0.57 are the standard errors of the intercept and slope, respectively.

The degrees of freedom for the t-test are 29 + 81 - 2 = 108.

At the 5% level of significance, the critical value for a one-tailed t-test with 108 degrees of freedom is -1.66.

Since We can conclude that females, on average, are shorter than males in the population from which the sample was drawn.

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The critical value of t is approximately 1.895.

The critical value of t is approximately 2.626.

What is Confidence Interval?

In statistics, a confidence interval is a range of values calculated from a sample of data that is likely to contain the true value of an unknown population parameter with a certain level of confidence, usually expressed as a percentage. It is a measure of the precision and reliability of an estimate.

a) Using a t-distribution calculator or a t-table with 7 degrees of freedom and a 90% confidence level, the critical value of t is approximately 1.895.

b) Using a t-distribution calculator or a t-table with 103 degrees of freedom and a 99% confidence level, the critical value of t is approximately 2.626.

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Answer: the eastbound train has traveled 48 kilometers.

Step-by-step explanation: Let’s solve this problem. We can imagine the two trains starting at the origin of a coordinate plane, with the northbound train traveling along the y-axis and the eastbound train traveling along the x-axis. The northbound train has traveled 64 kilometers, so its position is (0, 64). The eastbound train has traveled some distance x along the x-axis, so its position is (x, 0).

The straight-line distance between the two trains is given by the distance formula: d = sqrt((x2 - x1)^2 + (y2 - y1)^2). Plugging in the coordinates of the two trains and the given distance of 80 kilometers, we get: 80 = sqrt((x - 0)^2 + (0 - 64)^2). Squaring both sides and simplifying, we get: 6400 = x^2 + 4096. Solving for x, we get: x^2 = 2304. Taking the square root of both sides, we get: x = sqrt(2304) = 48.

So, the eastbound train has traveled 48 kilometers.