What is the height of the models sail in centimeters?

Answer:

0.85

Step-by-step explanation:

Answer:

Step-by-step explanation:

The three ways to write a ratio is:

8/4

8 to 4

8:4

The ratio in simpliest form is 2/1

You would divide each "side" by 4

Answer: 3x-x+2=4

Step-by-step explanation:

12/X=3/6 CROSS MULTIPLY.

3X=12*6

3X=72 X=72/3 X= 24 FEET TALL.

we are 99% confident that the true value of µ lies within the interval (12.5831, 13.9169).

We have a random sample from a normal distribution with mean µ and a true variance of 0.81.

From this sample of 16 values, the sample mean was 13.25.

We want to calculate the 99% confidence interval for µ.

We can use the t-distribution to calculate the confidence interval since the sample size is less than 30.

We need to find the t-value that corresponds to a 99% confidence interval and 15 degrees of freedom (n-1).

We can use a t-distribution table or calculator to find that the t-value is 2.9477.

Using this value, we can calculate the confidence interval as follows: Lower bound = sample mean - (t-value * standard error)Upper bound = sample mean + (t-value * standard error) The standard error is the standard deviation divided by the square root of the sample size.

So, in this case: Standard error = √(0.81/16) = 0.2025 Lower bound = 13.25 - (2.9477 * 0.2025) = 12.5831Upper bound = 13.25 + (2.9477 * 0.2025) = 13.9169Therefore, the 99% confidence interval for µ is (12.5831, 13.9169).

This means that if we repeated the process of taking a sample of 16 values many times and calculating a confidence interval each time, we would expect that 99% of those intervals would contain the true value of µ.

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Given: Sample size, n = 16, Sample mean, x = 13.25, Population variance, σ² = 0.81, Confidence level = 99%. The 99% confidence interval for the population mean, µ, is (12.676, 13.824). It means that if we repeat the process of drawing samples of size 16 from the same population infinite times and calculate the confidence intervals, then 99% of those confidence intervals will contain the true population mean.

Since the sample size is greater than 30 and we have a known population variance, we can use the z-distribution for finding the confidence interval for the population mean.

We can use the following formula to find the confidence interval at a given confidence level.

x - z(α/2) * σ/√n < µ < x + z(α/2) * σ/√n, Where z(α/2) is the z-value at α/2 level of significance.

z(α/2) can be found from the standard normal distribution table.

At 99% confidence level,

α = 1 - 0.99

= 0.01.

α/2 = 0.01/2

= 0.005.

At α/2 = 0.005 level of significance,

z(α/2) = 2.576

σ = √0.81

= 0.9

Substituting the values in the formula,

x - z(α/2) * σ/√n < µ < x + z(α/2) * σ/√n

13.25 - 2.576 * 0.9/√16 < µ < 13.25 + 2.576 * 0.9/√16

13.25 - 0.574 < µ < 13.25 + 0.57412.676 < µ < 13.824

Interpretation of Interval: The 99% confidence interval for the population mean, µ, is (12.676, 13.824).

It means that if we repeat the process of drawing samples of size 16 from the same population infinite times and calculate the confidence intervals, then 99% of those confidence intervals will contain the true population mean.

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Answer:

696.348

Step-by-step explanation:

Answer:

14.5

Step-by-step explanation:

Slope is the change in y over the change in x

1. (1,1) and (4,5)

Slope = (5-1)/(4-1)

Slope = 4/3

2. (2,-1) and (4,13)

Slope = (13 - -1)/ (4-2)

Slope = 14/2

Slope = 7

3. (1,18) and (-9,-2)

Slope = (-2 -18)/(-9 -1)

Slope = -20/-10

Slope = 2

Answer:

6

10

Step-by-step explanation:

30 5 6

70 7 10

Answer:

x =5, y = 4

Step-by-step explanation:

3x + 4y = 31..... (1)

2x - 4y = -6..... (2)

Adding equations (1) & (2)

[tex]3x + \cancel{4y} = 31 \\2x - \cancel{4y} = -6\\ - - - - - - - \\ 5x = 25 \\ x = \frac{25}{5} \\ \bold{ \purple{x = 5}} \\ plug \: x = 5 \: in \: eq \: (1) \\ 3(5) + 4y = 31 \\ 15 + 4y = 31 \\ 4y = 31 - 15 \\ 4y = 16 \\ y = \frac{16}{4} \\ \bold{ \red{y = 4}}[/tex]

Answer: No image? Incomplete question? FREE POINTS!

Step-by-step explanation:

BRAINLIEST PLEASE!

Answer:

D. the average difference in height between each player and the mean height

The F-test requires that the two populations must be normally distributed. Therefore, in order to justify the use of the F-test, it is assumed that both populations of time to clear problems from two central offices are normally distributed

a. At the 0.05 level of significance, is there evidence of a difference in the variability of the time to clear problems between the two central offices?

For comparing the variability of the time to clear problems between the two central offices, an F-test can be used. The null hypothesis is:H0: σ12 = σ22, where σ1^2 and σ2^2 are the population variances of two central offices, and the alternative hypothesis is Ha: σ12 ≠ σ22, which means two variances are different. For this study, the significance level is 0.05. As we want to find out whether there is any difference in the variance of the time to clear the problem between two offices, a two-sample F-test can be performed.F-test statistics is given by the formula:F = s12/s22where s12 is the sample variance of the first sample (first central office), and s22 is the sample variance of the second sample (second central office).We can use Excel to calculate the F statistic.Using the given dataset, the F statistic is calculated as: σ12 = 22.66666667, σ22 = 25.25, F = 0.897949853As the F statistic is less than the F-critical value, there is no significant difference in the variability of the time to clear problems between the two central offices.b. Interpret the p-value.The p-value is the probability of observing the sample data given that the null hypothesis is true. If the p-value is less than the level of significance (α = 0.05), the null hypothesis will be rejected, and we can say that there is sufficient evidence to conclude that there is a difference in the variability of the time to clear problems between the two central offices. The p-value of this F-test is 0.467. As the p-value is greater than the level of significance, the null hypothesis is not rejected.c. What assumption do you need to make in (a) about the two populations in order to justify your use of the F test?

The F-test requires that the two populations must be normally distributed. Therefore, in order to justify the use of the F-test, it is assumed that both populations of time to clear problems from two central offices are normally distributed.

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The data set contains samples of 20 problems.

Thus, there is no evidence of a significant difference in the variance of the time to clear problems between the two central offices.

The p-value is 0.17.

Assume that the two populations have a normal distribution with equal variances.

a. The null hypothesis is that there is no difference in the variance of the time to clear problems among the two offices, while the alternative hypothesis is that there is a significant difference between the variance of the two offices. Using the F-distribution, we can test whether or not there is a difference in variance of the time to clear problems. The formula for F-value is given below:

F-value = s1^2 / s2^2

Where s1^2 and s2^2 are the variances of the two samples. With the help of the provided data, we can calculate the variances for the two samples, which are as follows:

s1^2 = 42.08

s2^2 = 22.80

Then, we can calculate the F-value as follows:

F = s1^2 / s2^2

= 42.08 / 22.80

= 1.84

Using the F-distribution table, we can find the critical value of F as 2.17 (with 19 degrees of freedom for both the numerator and denominator).Since the calculated F-value (1.84) is less than the critical value of F (2.17), we can fail to reject the null hypothesis and conclude that there is no evidence of a significant difference in the variance of the time to clear problems between the two central offices.

b. The p-value represents the probability of observing a test statistic as extreme as the one calculated, assuming that the null hypothesis is true. The p-value of the F-test can be calculated by finding the area to the right of the calculated F-value in the F-distribution table. In this case, the p-value is 0.17 (using a two-tailed test).

c. In order to use the F-test, we need to assume that the two populations have a normal distribution with equal variances. Furthermore, the samples should be independent and randomly selected. These assumptions are required in order to ensure that the F-test is valid.

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A.) Similar?  YES  

Why or why not?  It is similar because as your can see in the picture above, it shows a large triangle and a small triangle. On the smaller triangle, the corner "G" has a degree of 67. On the larger triangle, the corner "P" has the same degree as the smaller one, therefore all the other corners have the same degree making the missing degree for "R" 34 degrees, and the missing "C" 79 degrees.  

If so, *similarity statement and scale factor: SF:  1.5  

*I didn't know what they meant by similarity statement

B.) Similar?  NO  

Why or why not?  The triangles are two totally different shapes, with totally different degree angles, making it totally different  

If so, *similarity statement and scale factor: SF:  2  

*I didn't know what they meant by similarity statement

C.) Similar?  NO  

Why or why not?  The triangles are two totally different shapes, with totally different degree angles, making it totally different  

If so, *similarity statement and scale factor: SF:  1.2  

*I didn't know what they meant by similarity statement

D.) Similar?  YES  

Why or why not?  They are the same shape; just with one enlarged and one decreased in size. They both have the same angle degree and everything else similar but the size of the shape.  

If so, *similarity statement and scale factor: SF:  2  

*I didn't know what they meant by similarity statement

**To find the scale factor you would have to divide the big number, to the smaller number... Like for figure D, the base of the BIG triangle is 56, and the small triangle is 28. 56 ÷ 28 = 2 Making 2 the SF (Scale Factor)

Answer:

so you multiply 12 by 8 =96

Answer:

area of large circle=pi×r²

area of large circle =3.14×12²

area of large circle=452.16in²

area of small circle=pi×r²

area of small circle=3.14×8²

area pf small circle=200.96in²

area of shaded circle=452.16in²- 200.96in²

area of shaded circle =251.2in²

Answer:

I answered it on the second time you asked this question check it out

Answer:

c. 60

Step-by-step explanation:

Given

[tex]n = 61[/tex] --- sample

[tex]\bar x = 23[/tex]

Required

Determine the degrees of freedom (df)

This is calculated as:

[tex]df = n - 1[/tex]

[tex]df = 61 - 1[/tex]

[tex]df = 60[/tex]

Answer:

A≈176.71

Step-by-step explanation:

A=πr^2

We find the radius by slicing the diameter in half. The radius is half of the diameter.

A = A=π 7.5^2

A=π 56.25

A≈176.71

Answer:

V = 36 cm³

Step-by-step explanation:

V = 1/2(3)(4)(6) = 36 cm³

it's recorded that out of 1000 people, 762 wear the corrective lenses.

just divide 762 from 1000 and multiply that result by 100.

762/ 1000 = .762

.762 x 100 = ? %

which is 76.2 %

so, we predict that 76.2% of Americans would wear corrective lenses.

Answer:

D

Step-by-step explanation:

6x7x4= 168

The given differential equation by undetermined coefficients is y'' - 10y' + 25y = 30x + 3. Its solution is as follows: Let us assume y = yh + yp where yh is the homogeneous solution and yp is the particular solution. To find the homogeneous solution, solve the following differential equation: y'' - 10y' + 25y = 0characteristic equation: r2 - 10r + 25 = 0(r - 5)2 = 0Thus, yh = c1e5x + c2xe5x

Now, let us find the particular solution by assuming the following particular solution: yp = Ax + B Substituting this into the differential equation: y'' - 10y' + 25y = 30x + 3 yields:-10A + 25B = 3, and0x + A = 30Solving for A and B: A = 30 and B = 15Thus, yp = 30x + 15Therefore, the general solution is:y = yh + yp = c1e5x + c2xe5x + 30x + 15, where c1 and c2 are arbitrary constants.

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Answer:

1/2

Step-by-step explanation:

Hope it helps :)

a) Var[M9(X)] = 25/9.

b) P[X > 7] = exp(-7/5).

c) P[Z < 2/5] = 0.3446.

Given information: X1, . . . , Xn is an iid sequence of exponential random variables, each with expected value 5.

(a) We know that the sample mean based on nine trials is M9(X). Now, to calculate the variance of the sample mean based on nine trials, Var[M9(X)], we can use the formula for the variance of a sample mean, which is:

Var[M9(X)] = Var[X]/9  .

Since X is an exponential random variable with expected value 5, its variance is 5^2 = 25. Thus,Var[M9(X)] = 25/9.

(b) To find P[X * 1 > 7], we can use the probability density function of an exponential distribution, which is given by:

f(x) = 1/5 exp(-x/5), x > 0  .

Now, using this probability density function, we have:

P[X > 7] = ∫7∞f(x) dx= ∫7∞ 1/5 exp(-x/5) dx.

Using integration by substitution, with u = x/5 and du = (1/5)dx, we have:

P[X > 7] = ∫7/5∞ exp(-u) du= exp(-7/5).

(c) Since we know that X1, . . . , Xn is an iid sequence of exponential random variables, each with expected value 5 and variance 25, we can apply the central limit theorem. According to the central limit theorem, the sample mean M9(X) is approximately normally distributed with mean 5 and variance 25/9. Thus, we have:

P[M9(X) > 7] = P[Z > (7-5)/(5/3)] where Z is a standard normal random variable. This simplifies to:

P[M9(X) > 7] = P[Z > 2/5] = 1 - P[Z < 2/5]

Using the standard normal distribution table or calculator, we get:

P[Z < 2/5] = 0.6554P[M9(X) > 7] = 1 - P[Z < 2/5] = 1 - 0.6554 = 0.3446.

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Answer:

The statement that best describes how the author of "The Train to Somewhere" alters historical details to tell her story is option B;

B. The author describes the experiences and feelings of an imaginary orphan who was sent west on an orphan train

Step-by-step explanation:

The author included the experiences of one of the children who according to the historical details in the historical account of the Orphan Train Riders were aided by the minister, Charles Loring Brace, to be taken from the New York City streets on the East to other states and Canada to the West

The child, a 5 year old girl, who had lost both parents had been living on the street with her brother for some months

The author also highlighted the taught by the child, of being separated from her brother at one of the destination towns of the train, and the manifestation of the beliefs of the minister, Brace, of the children finding loving families that would help them become productive members of society.

The values of b = 0 or a = -3 -  such that zw is real, letting z = 3+ bi and w = a + bi where a, b E R.

To determine the value of b in terms of a such that zw is real, we first need to find zw. Using the distributive property, we have:

zw = (3 + bi)(a + bi)

zw = 3a + 3bi + abi - b^2

To make zw real, the imaginary part must be equal to zero. Therefore, we have:

3b + ab = 0

b(3 + a) = 0

Since b cannot be equal to zero (otherwise z and w would be real), we have:

a = -3

Therefore, b = 0 or a = -3 - this is the value of a such that zw is real.

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Two people are trying to decide whether a die is fair. The correct test for analyzing the fairness of the die with the given data is the chi-square [tex]X^2[/tex] test, not the z-test.

The z-test is used for analyzing data when we have known population parameters, such as the mean and standard deviation. However, in this case, we are dealing with categorical data (the frequencies of each face of the die), and we want to determine if the observed frequencies significantly differ from the expected frequencies.

To perform the chi-square test, we first need to calculate the expected frequency for each face of the die. The expected frequency is calculated by multiplying the total number of rolls (100) by the probability of each face (1/6, assuming a fair die). Each face of the die is expected to occur approximately 16.67 times (100/6 = 16.67).

Next, we calculate the degrees of freedom (df) for the chi-square test. For a fair die with 6 faces, the df is (number of categories - 1), which is 5 in this case.

Then, we calculate the chi-square statistic[tex](X^2)[/tex] by summing the squared differences between the observed and expected frequencies, divided by the expected frequencies. The [tex]X^2[/tex] value is used to assess the goodness-of-fit between the observed and expected frequencies.

Finally, we determine the p-value associated with the calculated [tex]X^2[/tex]value using the chi-square distribution and the degrees of freedom. The p-value indicates the likelihood of observing the data if the die is fair.

To provide the specific values for the expected frequency, degrees of freedom, [tex]X^2[/tex], and p-value, the actual calculations based on the given data are required.

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Answer:

13 in

Step-by-step explanation:

Let s be the length of a side of the triangle

then one rectangular face of the triangular prism

would be s x 14.

There are 3 rectangular faces in the triangular prism so

Lateral Surface Area  = 3 x (s x 14)

182 = 42s

182/42 = s

There are 3 sides to the triangle so

3 x (182/42) = 13