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3 years ago

The local school board needs to choose whether to build an addition to the elementary school library or remodel the

Mathematics

1 answer:

11Alexandr11 [23.1K]3 years ago

7 0

Answer:

A random sample of 10% of all taxplayers

Step-by-step explanation:

Smaller sample sizes may not be representive of the population

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Solve the inequality for u. 24 Simplify your answer as much as possible.

Annette [7]

Answer:

hey i think you forgot to add the expression, would u mind adding the question :) thanks

Step-by-step explanation:

3 0

3 years ago

(a) Find the size of each of two samples (assume that they are of equal size) needed to estimate the difference between the prop

zalisa [80]

Answer:

(a) The sample sizes are 6787.

(b) The sample sizes are 6666.

Step-by-step explanation:

(a)

The information provided is:

Confidence level = 98%

MOE = 0.02

n₁ = n₂ = n

$\hat p_{1} = \hat p_{2} = \hat p = 0.50\ (\text{Assume})$

Compute the sample sizes as follows:

$MOE=z_{\alpha/2}\times\sqrt{\frac{2\times\hat p(1-\hat p)}{n}$

$n=\frac{2\times\hat p(1-\hat p)\times (z_{\alpha/2})^{2}}{MOE^{2}}$

$=\frac{2\times0.50(1-0.50)\times (2.33)^{2}}{0.02^{2}}\\\\=6786.125\\\\\approx 6787$

Thus, the sample sizes are 6787.

(b)

Now it is provided that:

$\hat p_{1}=0.45\\\hat p_{2}=0.58$

Compute the sample size as follows:

$MOE=z_{\alpha/2}\times\sqrt{\frac{\hat p_{1}(1-\hat p_{1})+\hat p_{2}(1-\hat p_{2})}{n}$

$n=\frac{(z_{\alpha/2})^{2}\times [\hat p_{1}(1-\hat p_{1})+\hat p_{2}(1-\hat p_{2})]}{MOE^{2}}$

$=\frac{2.33^{2}\times [0.45(1-0.45)+0.58(1-0.58)]}{0.02^{2}}\\\\=6665.331975\\\\\approx 6666$

Thus, the sample sizes are 6666.

7 0

2 years ago

−8x= \,\,7-(7.2x+8.8)+x 7−(7.2x+8.8)+x

katrin [286]

Answer:

math was made just so we can be bored

Step-by-step explanation:

7 0

3 years ago

Jack use the place value chart below to show 1,826.5. Jack divided 1,826.5 by a number to get a quotient of 18.265 which he wrot

Annette [7]

Answer:

Step-by-step explanation:

Let the number we are looking for be x. If Jack divided 1,826.5 by the number to get a quotient of 18.265, to get the number we will solve the equation;

1,826.5/x = 18.265

Cross multiply

1826.5 = 18.265x

Divide both sides by 18.265

18.265x = 1826.5

18.265x/18.625 = 1826.5/18.625

x = 100

Hence Jack divide 1,826.5 by 100 to get a quotient of 18.265

3 0

3 years ago

The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e

I am Lyosha [343]

Answer:

(a) The proportion of tenth graders reading at or below the eighth grade level is 0.1673.

(b) The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.198, 0.260).

Step-by-step explanation:

Let X = number of students who read above the eighth grade level.

(a)

A sample of n = 269 students are selected. Of these 269 students, X = 224 students who can read above the eighth grade level.

Compute the proportion of students who can read above the eighth grade level as follows:

$\hat p=\frac{X}{n}=\frac{224}{269}=0.8327$

The proportion of students who can read above the eighth grade level is 0.8327.

Compute the proportion of tenth graders reading at or below the eighth grade level as follows:

$1-\hat p=1-0.8327$

$=0.1673$

Thus, the proportion of tenth graders reading at or below the eighth grade level is 0.1673.

(b)

the information provided is:

n = 709

X = 546

Compute the sample proportion of tenth graders reading at or below the eighth grade level as follows:

$\hat q=1-\hat p$

$=1-\frac{X}{n}$

$=1-\frac{546}{709}$

$=0.2299\\\approx 0.229$

The critical value of z for 95% confidence interval is:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

Compute the 95% confidence interval for the population proportion as follows:

$CI=\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

$=0.229\pm 1.96\times \sqrt{\frac{0.229(1-0.229)}{709}}\\=0.229\pm 0.03136\\=(0.19764, 0.26036)\\\approx (0.198, 0.260)$

Thus, the 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.198, 0.260).

5 0

3 years ago