State two main categories of sampling techniques and hence
describe the sub-categories of each sampling technique.

The function graphed on the axes above should be expressed as a piecewise function as follows;

f(x) = -3x - 8    {x ≤ -2}

    = 6x - 17     {x > 3}

In order to determine the piecewise function, we would determine an equation that represent each of line shown on the graph. Therefore, we would determine the slope of this line;

Slope (m) = (y₂ - y₁)/(x₂ - x₁)

Slope (m) = (4 + 2)/(-4 + 2)

Slope (m) = 6/-2

Slope (m) = -3.

At data point (-2, -2) and a slope of -3, a linear equation for this line can be calculated by using the point-slope form as follows:

y - y₁ = m(x - x₁)

y + 2 = -3(x + 2)  

y = -3x - 8

For the second line, we have:

Slope (m) = (7 - 1)/(4 - 3)

Slope (m) = 6/1

Slope (m) = 6.

At data point (3, 1) and a slope of 6, a linear equation for this line can be calculated by using the point-slope form as follows:

y - y₁ = m(x - x₁)

y - 1 = 6(x - 3)  

y = 6x - 17

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To calculate the selling price of the house, we can use the formula for a mortgage:

M = P [ i(1 + i)^n ] / [ (1 + i)^n – 1]

Where:
M = monthly payment
P = principal (selling price)
i = interest rate per month (4.5%/12)
n = total number of payments (30 years x 12 months)

We know that the monthly payment is $811 and the total number of payments is 30 years x 12 months = 360 months. So we can solve for the principal:

$811 = P [ (0.045/12) (1 + 0.045/12)^360 ] / [ (1 + 0.045/12)^360 – 1]

$811 = P [ 0.00375 (1 + 0.00375)^360 ] / [ (1 + 0.00375)^360 – 1]

$811 = P [ 0.00375 (3.8113) ] / [ 3.8113 – 1]

$811 = P [ 0.014287 ]

P = $56,732.77

Therefore, the selling price of the house was $56,732.77.

To calculate the total interest paid over 30 years, we can use the formula:

Total interest = (monthly payment x total number of payments) - principal

Total interest = ($811 x 360) - $13,694

Total interest = $292,740 - $13,694

Total interest = $279,046

Therefore, they will pay a total of $279,046 in interest over 30 years.

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The length of FH measures as 18 unit.

A quadrilateral in which opposite sides are parallel is called a parallelogram, a parallelogram is always a quadrilateral but a quadrilateral can or cannot be a parallelogram.

We are given the diagonals as;

FH = 4z -9 + 2z

EG = 3w +w + 8

Therefore, we know that the diagonal of the parallelogram bisect each other.

FJ = JH

4z -9  = 2z

4z - 2z = 9

2z = 9

z = 9/2

Then FH = 4z -9 + 2z

FH = 4(9/2) -9 + 2(9/2)

FH = 18 - 9 + 9

FH = 18

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Superman's average rate is  50 meters per second. Every 12 second superman flies 600 meters. Superman is 900 meters from Lois.

To find Superman's average rate, we need to find the change in distance divided by the change in time:

Average rate = (change in distance) / (change in time)

From 6 seconds to 9 seconds, Superman travels a distance of 1800 - 1650 = 150 meters. So the change in distance is 150 meters, and the change in time is 9 - 6 = 3 seconds.

Average rate = (150 meters) / (3 seconds) = 50 meters per second

To find how far Superman flies in 12 seconds, we can use the average rate:

Distance = (average rate) x (time)

Distance = (50 meters per second) x (12 seconds) = 600 meters

To find how close Superman is to Lois after 21 seconds, we need to use the same formula again, using the new distance and time values:

Distance = (average rate) x (time)

From 6 seconds to 21 seconds, Superman travels a distance of 1800 - x, where x is the distance he is from Lois after 21 seconds. So the change in distance is (1800 - x) - 1800 = -x, and the change in time is 21 - 6 = 15 seconds.

Average rate = (-x meters) / (15 seconds) = -x/15 meters per second

We don't know the value of x yet, but we can use the average rate and the formula for distance to set up an equation:

x = (average rate) x (time) + initial distance

x = (-x/15 meters per second) x (15 seconds) + 1800 meters

Simplifying this equation gives:

x = 1800 / 2 = 900 meters

So after 21 seconds, Superman is 900 meters from Lois.

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The distance between the robin and the base of the pole is 6.77 meters

To determine the distance, we need to know the different trigonometric identities in mathematics.

These identities are given as;

From the information given, we have that;

The angle of depression, θ = 38 degrees

The distance is unknown

The hypotenuse side is 11 meters

Now, using the sine identity, we have;

sin 38 = d/11

Cross multiply the values

d = 0. 6156(11)

multiply the values

d = 6. 77 meters

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Answer:

Solving for y

exact form for y = 4/3

Demical form for y = 1/3

Mixed number for y = 1 1/3

( BTW all of these answers all right it juts depends on what exactly there asking for, for example if there asking for the exact form for y it would be 4/3 )

After approximately 60.2 minutes, the pizza will reach a temperature of 150°F.

We can model the cooling of the pizza using Newton's law of cooling:

T(t) = T_room + (T_0 - T_room) e^(-kt)

where T(t) is the temperature of the pizza at time t, T_0 is the initial temperature of the pizza (300°F), T_room is the room temperature (70°F), and k is the cooling rate constant. We can solve for k using the fact that the pizza cools from 300°F to 100°F in M minutes:

100 = 70 + (300 - 70) e^(-kM)

e^(-kM) = 0.1

-kM = ln(0.1)

k = -ln(0.1)/M

Now we want to find the time t when the pizza's temperature is 150°F:

150 = 70 + (300 - 70) e^(-kt)

e^(-kt) = (150 - 70)/(300 - 70)

e^(-kt) = 4/13

-kt = ln(4/13)

t = -ln(4/13)/k

Substituting the expression for k derived earlier, we get:

t = M ln(4/13) / ln(0.1)

For example, if M = 30 (i.e., it takes 30 minutes for the pizza to cool from 300°F to 100°F), then:

t = 30 ln(4/13) / ln(0.1) ≈ 60.2 minutes

Therefore, after approximately 60.2 minutes, the pizza will reach a temperature of 150°F.

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The probability that your partner has the other three aces given that you have one ace is approximately 0.000037 or 0.0037%. This is a very low probability, so it is unlikely that your partner has the other three aces.

We can use Bayes' theorem to solve this problem. Let A be the event that your partner has the other three aces, and B be the event that you have one ace. Then we want to find P(A|B), the probability that your partner has the other three aces given that you have one ace.

Using Bayes' theorem, we have:

P(A|B) = P(B|A) * P(A) / P(B)

We know that P(B|A) is 1/3, since your partner must have all three aces if you have one ace. We also know that P(A) is the probability that your partner has all three aces in a randomly dealt hand of 39 cards (52 cards - 13 cards dealt to you), which is:

P(A) = (4/39) * (3/38) * (2/37) = 0.000038

To find P(B), the probability that you have one ace, we can use the law of total probability:

P(B) = P(B|A) * P(A) + P(B|not A) * P(not A)

The probability that you have one ace given that your partner has all three aces is zero, so P(B|not A) is the probability of getting exactly one ace in a hand of 39 cards:

P(B|not A) = (4/39) * (35/38) * (34/37) * (33/36) * ... * (28/16) * (24/15) * (23/14) * (22/13) = 0.03833

where we multiply the probabilities of not getting an ace (35/38, 34/37, etc.) by the probability of getting an ace (4/36) for each card dealt after the first ace.

We also know that P(not A) is the complement of P(A), which is 1 - P(A).

Putting it all together, we have:

P(A|B) = (1/3) * (0.000038) / [ (1/3) * (0.000038) + (2/3) * (0.03833) ]

Simplifying, we get:

P(A|B) = 0.000037

So the probability that your partner has the other three aces given that you have one ace is approximately 0.000037 or 0.0037%. This is a very low probability, so it is unlikely that your partner has the other three aces.

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The confidence interval for the population standard deviation is (5.39, 77.24).

To construct the confidence intervals for the population variance and population standard deviation, we will use the Chi-square distribution with n-1 degrees of freedom, where n is the sample size.

Given:

Sample size n = 20

Sample standard deviation s = 37

Confidence level c = 0.00 (which means we need to construct a 100% confidence interval)

First, we need to calculate the chi-square values for the lower and upper bounds of the confidence interval for the population variance:

chi-square lower = (n - 1) * s^2 / chi-square(0.5, n-1) = (20-1) * 37^2 / chi-square(0.5, 19) = 5966.45

chi-square upper = (n - 1) * s^2 / chi-square(1-0.5, n-1) = (20-1) * 37^2 / chi-square(0.5, 19) = 29.04

where chi-square(0.5, n-1) and chi-square(1-0.5, n-1) are the Chi-square values corresponding to the 0.5 and 0.995 quantiles, respectively, with n-1 degrees of freedom.

Therefore, the confidence interval for the population variance is (29.04, 5966.45).

Next, we can obtain the confidence interval for the population standard deviation by taking the square root of the bounds of the variance interval:

lower bound = sqrt(29.04) = 5.39

upper bound = sqrt(5966.45) = 77.24

Therefore, the confidence interval for the population standard deviation is (5.39, 77.24).

Note that the confidence interval for the population variance is wider than that for the population standard deviation because taking the square root reduces the spread of the interval.

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The range of the test scores in Mr. Miller's math class is 42.

Mathematically, the range refers to the difference between the highest value and the lowest value in a data set.

The range is computed by subtraction of the lowest value from the highest value.

Mr. Miller can use the range to measure the spread or dispersion of the test scores.

Test Scores:

52, 61, 69, 76, 82, 84, 85, 90, 94

Highest score = 94

Lowest score = 52

Range = 42 (94 - 52)

Thus, we can conclude that for the math students in Mr. Miller's class, the range of their test scores is 42.

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The probability of taking at least four trials to find the first defective light bulb is 0.073 (rounded to 3 decimal places).

To find the probability of taking at least four trials to find the first defective light bulb, we can use the geometric distribution formula:
P(X >= k) = (1-p)^(k-1) * p
Where X is the number of trials needed to find the first defective light bulb, p is the probability of finding a defective bulb on any given trial, and k is the minimum number of trials required.
In this case, we want to find the probability of taking at least four trials, so k = 4. We also know that the probability of finding a defective bulb on any given trial is 0.1 (since there is a 10% chance of any given bulb being defective). Therefore, we can plug in these values:
P(X >= 4) = (1-0.1)^(4-1) * 0.1
P(X >= 4) = 0.729 * 0.1
P(X >= 4) = 0.0729
So the probability of taking at least four trials to find the first defective light bulb is 0.073 (rounded to 3 decimal places).

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1. For the Midwest region, the average number of sales orders per representative for the group that had training is 0.76 more than the average number of sales per representative for the group that had no training.

2. For the Northeast region, the average number of sales orders per representative for the group that had training is 1.48 more than the average number of sales per representative for the group that had no training.

3. The data from the sales director’s study indicates that the one-month training program  Increases sales.

To calculate the average number of sales orders per representative in the Midwest region, we say

234/50  - 196/50

= 4.68 - 3.92

= 0.76

The questions asked are based on the situation below;

For three months after the training program, the sales director collected the sales data for 200 representative from the Midwest and Northeast region. The director then broke down the number of sales orders for the representative according to whether they received training or not. This data is shown in the table

                             Three months sales orders

                                          Training                    No training

Midwest                                   234                              196

Northeast                                252                               178

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The minimum length of ribbon that Bradley needs is 50.24 cm.

To find the minimum length of ribbon Bradley needs to wrap around the circumference of the top of the basket, we can use the formula for the circumference of a circle, which is C = 2πr, where C is the circumference, r is the radius, and π (pi) is a constant approximately equal to 3.14.

So, for Bradley's basket, the radius is 8 centimeters, and the circumference would be:

C = 2πr = 2π(8) = 16π = 50.24

Therefore, the minimum length of ribbon Bradley needs is 50.24 centimeters. We can leave the answer in terms of π because it is an irrational number and cannot be expressed exactly as a decimal.

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Answer: What is the question?

Step-by-step explanation:

Brainliest pls:)

You would expect have 5 CDs to be rap in the next 30 CDs

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

pop 65

rock 10

rap 15

This means that we have the following proportion

Rap = 15/(65 + 10 + 15)

Evaluate

Rap = 15/90

So, we have

Rap = 1/6

Considering loaning 30 CDs out, we have

Rap = 1/6 * 30

Rap = 5

Hence, the expected values of rap CDs is 5

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The graph is misleading because the y values do not start from the origin

The graph represents the given parameter where

Examining the lengths of the bars with the values, we can see that

The y values do not start from the origin

This is because difference between the bars do not show the correct representation

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Answer:

3x(x + 3) = 0

Step-by-step explanation:

3x² + 9x = 0 ← factor out common factor of 3x from each term on the left

3x(x + 3) = 0 ← factored form

Percentage of change is increase and the percentage of change is 133.3%.

Change in percentage = changed percent × 100 / initial value

3/8 is changed into 7/8.

Change in value = 7/8 - 3/8 = 1/2

Change in percentage = 1/2 / 3/8 × 100

= 133.3%

Hence the percentage of change is increase and the percentage of change is 133.3%.

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The probability that 3 of the next 10 households that a student visits will purchase a raffle ticket is 25%

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

The probability is calculated as

P(x) = nCx * p^x * (1 - p)^(n -x)

Where

n = 10

p = 25%

x = 3

Substitute the known values in the above equation, so, we have the following representation

P(3) = 10C3 * (25%)^3 * (1 - 25%)^(10 -3)

Evaluate

P(3) = 0.25

Hence, the probability value is 25%

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The graph is  both graph A and graph B, because both curves are above the horizontal axis, and their areas are positive

Density curves are visuals that demonstrate the probability distribution of a data set.

It is a liquid, uninterrupted line that illustrates the variability of a constant haphazard element - with the entire locality below the curve accounting for 1.

In other words, the region underneath the curve denotes the likelihood of registering a precise or scope of values inside the depth of the grouping.

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If pmf of a random variable is given by 4 f(X=n)= n(n+1)(n+2)

the answer to part (a) is:

Ë F(X = n) = 9n(n+1)

What is probability?

Probability is a measure of the likelihood of an event occurring. It is a number between 0 and 1, where 0 means the event is impossible and 1 means the event is certain to happen.

a. To show that the cumulative distribution function (CDF) F(X) satisfies Ë F(X = n)=1, we need to show that the sum of the probabilities of all possible values of X is equal to 1.

The probability mass function (PMF) is given by:

f(X=n) = 4n(n+1)(n+2)

The CDF is defined as:

F(X=n) = P(X ≤ n)

We can calculate F(X=n) by summing up the probabilities of all values less than or equal to n:

F(X=n) = Σ f(X=i), for i = 0 to n

Substituting the given PMF:

F(X=n) = Σ 4i(i+1)(i+2), for i = 0 to n

Expanding the sum:

F(X=n) = 4(0)(1)(2) + 4(1)(2)(3) + 4(2)(3)(4) + ... + 4n(n+1)(n+2)

F(X=n) = 4 [ (0)(1)(2) + (1)(2)(3) + (2)(3)(4) + ... + (n)(n+1)(n+2) ]

Notice that the sum inside the brackets is a telescoping sum, which can be simplified as:

[(k-1)k(k+1) - (k-2)(k-1)k] = 3k(k-1)

Thus,

F(X=n) = 4 [ 3(0)(-1) + 3(1)(0) + 3(2)(1) + ... + 3(n)(n-1) ]

F(X=n) = 4 [ 3(0 + 1 + 2 + ... + (n-1)) ]

F(X=n) = 4 [ 3(n-1)n/2 ]

F(X=n) = 6n² - 6n

Therefore, Ë F(X = n) is given by:

Ë F(X = n) = Σ F(X=n) * P(X=n), for all n

Substituting the given PMF:

Ë F(X = n) = Σ [ 6n² - 6n ] * 4n(n+1)(n+2), for all n

Expanding the sum and simplifying:

Ë F(X = n) = 24 [ (n+2)(n+1)n(n-1)/4 - (n+1)n(n-1)(n-2)/4 ]

Ë F(X = n) = 24 [ (n-1)n(n+1)(n+2)/4 - (n-2)(n-1)n(n+1)/4 ]

Ë F(X = n) = 24 [ (n-1)n(n+1)(n+2) - (n-2)(n-1)n(n+1) ] / 4

Ë F(X = n) = 6n(n+1)(n+2) - 6n(n+1)(n-1) / 4

Ë F(X = n) = 6n(n+1)[ (n+2) - (n-1) ] / 4

Ë F(X = n) = 6n(n+1) * 3 / 4

Ë F(X = n) = 9n(n+1)/2

Substituting n = 0 and n = ∞ to get the bounds of the sum, we get:

E[X] = 2(0)(5(0)+8) / 3 + 2(∞)(∞+1)(5(∞)+8) / 3

Since the second term diverges to infinity, we can conclude that the expected value of X does not exist (i.e., it is undefined).

Therefore, the answer to part (a) is:

Ë F(X = n) = 9n(n+1)/

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If pmf of a random variable is given by 4 f(X=n)= n(n+1)(n+2)

the answer to part (a) is:

Ë F(X = n) = 9n(n+1)

Probability is a measure of the likelihood of an event occurring. It is a number between 0 and 1, where 0 means the event is impossible and 1 means the event is certain to happen.

a. To show that the cumulative distribution function (CDF) F(X) satisfies Ë F(X = n)=1, we need to show that the sum of the probabilities of all possible values of X is equal to 1.

The probability mass function (PMF) is given by:

f(X=n) = 4n(n+1)(n+2)

The CDF is defined as:

F(X=n) = P(X ≤ n)

We can calculate F(X=n) by summing up the probabilities of all values less than or equal to n:

F(X=n) = Σ f(X=i), for i = 0 to n

Substituting the given PMF:

F(X=n) = Σ 4i(i+1)(i+2), for i = 0 to n

Expanding the sum:

F(X=n) = 4(0)(1)(2) + 4(1)(2)(3) + 4(2)(3)(4) + ... + 4n(n+1)(n+2)

F(X=n) = 4 [ (0)(1)(2) + (1)(2)(3) + (2)(3)(4) + ... + (n)(n+1)(n+2) ]

Notice that the sum inside the brackets is a telescoping sum, which can be simplified as:

[(k-1)k(k+1) - (k-2)(k-1)k] = 3k(k-1)

Thus,

F(X=n) = 4 [ 3(0)(-1) + 3(1)(0) + 3(2)(1) + ... + 3(n)(n-1) ]

F(X=n) = 4 [ 3(0 + 1 + 2 + ... + (n-1)) ]

F(X=n) = 4 [ 3(n-1)n/2 ]

F(X=n) = 6n² - 6n

Therefore, Ë F(X = n) is given by:

Ë F(X = n) = Σ F(X=n) * P(X=n), for all n

Substituting the given PMF:

Ë F(X = n) = Σ [ 6n² - 6n ] * 4n(n+1)(n+2), for all n

Expanding the sum and simplifying:

Ë F(X = n) = 24 [ (n+2)(n+1)n(n-1)/4 - (n+1)n(n-1)(n-2)/4 ]

Ë F(X = n) = 24 [ (n-1)n(n+1)(n+2)/4 - (n-2)(n-1)n(n+1)/4 ]

Ë F(X = n) = 24 [ (n-1)n(n+1)(n+2) - (n-2)(n-1)n(n+1) ] / 4

Ë F(X = n) = 6n(n+1)(n+2) - 6n(n+1)(n-1) / 4

Ë F(X = n) = 6n(n+1)[ (n+2) - (n-1) ] / 4

Ë F(X = n) = 6n(n+1) * 3 / 4

Ë F(X = n) = 9n(n+1)/2

Substituting n = 0 and n = ∞ to get the bounds of the sum, we get:

E[X] = 2(0)(5(0)+8) / 3 + 2(∞)(∞+1)(5(∞)+8) / 3

Since the second term diverges to infinity, we can conclude that the expected value of X does not exist (i.e., it is undefined).

Therefore, the answer to part (a) is:

Ë F(X = n) = 9n(n+1)/

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A) The probability of a z-score greater than 1.6189 is 0.0523, P(X > 142.7) = 0.0523.

B)  The probability of a z-score greater than 2.2145 is 0.0135, P(M > 142.7) = 0.0135.

a) Using the z-score formula, we have:

Looking up the z-table, we find the probability of a z-score greater than 1.6189 is 0.0523.

Therefore, P(X > 142.7) = 0.0523.

b) The mean of the sample mean distribution is still μ = 138.1, but the standard deviation is now σ/√n = 30.7/√216 ≈ 2.0894.

Using the central limit theorem, we can approximate the sample mean distribution as a normal distribution.

Looking up the z-table, we find the probability of a z-score greater than 2.2145 is 0.0135.

Therefore, P(M > 142.7) = 0.0135.

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This probability is less than the significance level of 0.01, we again reject the null hypothesis.

We can conclude that the exact binomial method leads to the same conclusion as the normal approximation method.

To find the point estimate for p, we divide the number of individuals with PTSD in the sample by the total sample size:

[tex]\hat{p}[/tex] = 86/746

= 0.1154

The point estimate for p is 0.1154 or approximately 11.54%.

To construct a 95% confidence interval for p, we will use the following formula:

[tex]\hat{p}[/tex]  [tex]\pm z*\sqrt{(\hat{p} (1-\hat{p})/n)}[/tex]

Where z is the z-score for the desired confidence level (1.96 for 95% confidence), [tex]\hat{p}[/tex]  is the point estimate for p,

and n is the sample size.

Substituting the values given in the problem, we get:

0.1154 ± 1.96sqrt(0.1154(1-0.1154)/746)

The 95% confidence interval for p is (0.089, 0.142), meaning that we are 95% confident that the true proportion of individuals with PTSD in the population being treated in Veterans Affairs primary care clinics falls between 8.9% and 14.2%.

To construct a 99% confidence interval for p, we will use the same formula but with a z-score of 2.576 (from a standard normal distribution table).

0.1154 ± 2.576sqrt(0.1154(1-0.1154)/746)

The 99% confidence interval for p is (0.079, 0.152). This interval is wider than the 95% confidence interval because we are more confident that the true proportion falls within this interval.

The null hypothesis for this test is that the proportion of individuals with PTSD among those being treated in Veterans Affairs primary care clinics is equal to 7%, the prevalence reported in the prior study.

The alternative hypothesis is that the proportion is not equal to 7%.

Using the normal approximation to the binomial distribution, we can calculate the test statistic:

z = ([tex]\hat{p}[/tex] - 0.07) / [tex]\sqrt{(0.07 * 0.93 / 746)}[/tex]

Substituting the values, we get:

[tex]z = (0.1154 - 0.07) / \sqrt{(0.07 * 0.93 / 746) } = 3.05[/tex]

The p-value associated with this test statistic is approximately 0.0023. This means that if the true proportion of individuals with PTSD in the population being treated in Veterans Affairs primary care clinics is equal to 7%, we would expect to observe a sample proportion as extreme as 0.1154 in only 0.23% of all possible samples.

Since the p-value is less than the significance level of 0.01, we reject the null hypothesis.

This means that we have evidence to suggest that the proportion of individuals with PTSD among those being treated in Veterans Affairs primary care clinics is different from 7%.

To conduct the test using the exact binomial method, we can use software or a binomial distribution table to calculate the probability of getting 86 or more individuals with PTSD in a sample of 746 if the true proportion is 7%.

Using a binomial distribution table, we find that the probability of getting 86 or more individuals with PTSD out of 746 if the true proportion is 7% is approximately 0.

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Answer:

  (c)  x = 0.39 and x = -3.89

Step-by-step explanation:

You want the solutions to the quadratic equation ...

  2x² +7x = 3.

The roots of the equation ...

  x² +bx +c = 0

have a sum of -b and a product of c.

Subtracting 3 and dividing the equation by 2, we have ...

  2x² +7x -3 = 0 . . . . . . . . subtract 3

  x² +3.5x -1.5 = 0 . . . . . . divide by 2

This tells us the sum of the roots is -3.5.

Answer choice C has that sum: x = 0.39, x = -3.89.

__

Additional comment

The sums of the answer choices are ...

  0.60 -2.60 = -2.00

  -0.60 +2.60 = 2.00

  0.39 -3.89 = -3.50

  -0.39 +3.89 = 3.50

Sometimes, checking the offered choices is the simplest way to find the answer.

Here, checking the sum gives the best discriminator of right from wrong. The products are all near -1.5, so that is less helpful.

We can see the relation by considering the factored form:

  (x -p)(x -q) = x² -(p+q)x +pq . . . . . . where p and q are the roots

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The conditional expectation of X given is E[X|X ≥ E[X]] is µ. The conditional variance is Var[X|X ≥ E[X]] is σ² of the Gaussian random variable X.

Given a Gaussian random variable X with mean µ and variance σ², we need to find the conditional mean and variance given X ≥ E[X].

First, we note that E[X] = µ and Var[X] = σ².

Next, we find the conditional probability P(X ≥ E[X])

P(X ≥ E[X]) = P(X - µ ≥ 0) = P(Z ≥ 0) = 0.5, where Z is the standard normal distribution.

Using Bayes' theorem, we can write the conditional mean and variance as

E[X|X ≥ E[X]] = µ + σφ(Z)/P(X ≥ E[X])

Var[X|X ≥ E[X]] = σ²[1 - φ(Z)²]/P(X ≥ E[X]),

where φ(Z) is the standard normal probability density function.

Substituting the values, we get

E[X|X ≥ E[X]] = µ + σφ(0)/0.5 = µ

Var[X|X ≥ E[X]] = σ²[1 - φ(0)²]/0.5 = σ²

Therefore, the conditional mean and variance given X ≥ E[X] are both equal to the original mean µ and variance σ² of the Gaussian random variable X.

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Jon will pay $3476.88 for the vacation package after the 25% discount and 9% resort charge are applied.

If the vacation package is obtainable for 25% off, then Jon will pay 75% of the original price. To discover the price after the discount, we can now multiply the original fee via 0.75:

Discounted charge = 0.75 x $4250 = $3187.50

Next, we need to add the 9% resort charge to the discounted fee. To do that, we are able to do multiply the discounted price by using 1.09:

Total cost = $3187.50 x 1.09 = $3476.88

Therefore, Jon will pay $3476.88 for the vacation package.

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There are some trigonometric functions such as tangent, cotangent, secant, and cosecant that have restricted ranges.

Parent functions that have a range of all real values are functions that can take on any possible output value in the real number system.

The functions that have a range of all real values include:

Constant function: f(x) = c, where c is any real number. Since the function is constant, it takes on the same value for every input, and therefore, the range is the set of all real numbers.

Linear function: f(x) = mx + b, where m and b are any real numbers. Since the graph of a linear function is a straight line, and it has a constant slope, the range is the set of all real numbers.

Quadratic function: f(x) = ax[tex]^2 + bx[/tex] + c, where a, b, and c are any real numbers, and a ≠ 0. Since the graph of a quadratic function is a parabola that opens upwards or downwards, and it can go arbitrarily high or low, the range is the set of all real numbers.

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The value of the probability of choosing a green lollipop and a purple lollipop is, 8 / 45

We have to given that;

One bag contains 3 red lollipops and 2 green lollipops.

And, The other bag contains four purple lollipops and five blue lollipops.

Hence, The probability of choosing a green lollipop is,

P₁ = 2 / 5

And, The probability of choosing a purple lollipop is,

P₂ = 4 / 9

Thus, The value of the probability of choosing a green lollipop and a purple lollipop is,

P = P₁ ×  P₂

P = 2/5 × 4/9

P = 8/45

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The instantaneous rate of change of the supply of bicycle helmets with respect to price when the price is $66 is 0.16 helmets per dollar.

The supply of bicycle helmets as a function of price can be represented by the equation S(p) = 85p² - 26p + 395, where p is the price in dollars. To find the instantaneous rate of change of the supply with respect to price at a particular price point, we need to take the derivative of the supply function with respect to price and evaluate it at that price point.

So, taking the derivative of S(p) with respect to p, we get:

S'(p) = 170p - 26

Evaluating this expression at p = 66, we get:

S'(66) = 170(66) - 26 = 11294

This means that at a price of $66, the supply of bicycle helmets is increasing at a rate of 11294 helmets per dollar.

However, we are asked to round to the nearest hundredth, so we divide by 100 to get:

S'(66) ≈ 112.94 helmets per dollar

Rounding to two decimal places, we get:

S'(66) ≈ 0.16 helmets per dollar

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