Let n be an integer in the conditional statement below.

According to the 2010 U.S. Census, there are approximately 8 million people living in Virginia. This is 2.8 million less than 4

galina1969 [7]

8 = 4N - 2.8
Add 2.8
10.8 = 4N
Divide 4
2.7 million is the population in Nevada

4 0

2 years ago

Read 2 more answers

A math teacher tells her students that eating a healthy breakfast on a test day will help their brain function and perform well

kogti [31]

Answer:

Step-by-step explanation:

Hello!

To test the claim that eating a healthy breakfast improves the performance of students on their test a math teacher randomly asked 46 students what did they have for breakfast before they took the final exam and classified them as:

<u>Group 1</u>: Ate healthy breakfast

X₁: Number of students that ate a healthy breakfast before the exam and earned 80% or higher.

n₁= 26

<u>Group 2: </u>Did not eat healthy breakfast

X₂: Number of students that did not eat a healthy breakfast before the exam and earned 80% or higher.

n₂= 20

After the test she counted the number of students that got 80% or more in the test for each group obtaining the following sample proportions:

p'₁= 0.50

p'₂= 0.40

The parameters of study are the population proportions, if the claim is true then p₁ > p₂

And you can determine the hypotheses as

H₀: p₁ ≤ p₂

H₁: p₁ > p₂

α: 0.05

$Z= \frac{(p'_1-p'_2)-(p_1-p_2)}{\sqrt{p'(1-p')[\frac{1}{n_1} +\frac{1}{n_2}] } } }$ ≈N(0;1)

pooled sample proportion: $p'= \frac{x_1+x_2}{n_1+n_2} =\frac{13+8}{46} = 0.46$

$Z_{H_0}= \frac{(0.5-0.4)-0}{\sqrt{0.46(1-0.46)[\frac{1}{26} +\frac{1}{20}] } } }= 0.67$

p-value: 0.2514

The decision rule is:

If p-value ≤ α, reject the null hypothesis.

If p-value > α, do not reject the null hypothesis.

The p-value: 0.2514 is greater than the significance level 0.05, the test is not significant.

At a 5% significance level you can conclude that the population proportion of math students that obtained at least 80% in the test and had a healthy breakfast is equal or less than the population proportion of math students that obtained at least 80% in the test and didn't have a healthy breakfast.

So having a healthy breakfast doesn't seem to improve the grades of students.

I hope this helps!

5 0

2 years ago

Zaire is buying a truck that costs p dollars. The expression shown represents the sales tax on the truck.

natima [27]

Answer:

The expression representing the total price of the truck including the sales tax is 1.07p

Step-by-step explanation:

The total price of the Truck can be calculated by adding the cost of the Truck to the sales tax of the Truck

From the question, the cost of the truck is $ p while the tax is 0.07p

So the total price of the truck will be;

p + 0.07p

= p( 1 + 0.07)

= p(1.07)

= 1.07p

8 0

2 years ago

From Statistics and Data Analysis from Elementary to Intermediate by Tamhane and Dunlop, pg 265. A thermostat used in an electri

Leviafan [203]

Answer:

$t=\frac{201.77-200}{\frac{2.41}{\sqrt{10}}}=2.32$

The degrees of freedom are given by:

$df=n-1=10-1=9$

The p value for this case is given by:

$p_v =2*P(t_{(9)}>2.32)=0.0455$

For this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly different from 200 F.

Step-by-step explanation:

Information given

data: 202.2 203.4 200.5 202.5 206.3 198.0 203.7 200.8 201.3 199.0

We can calculate the sample mean and deviation with the following formulas:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

$\sigma=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X=201.77$ represent the sample mean

$s=2.41$ represent the sample standard deviation

$n=10$ sample size

$\mu_o =200$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic

$p_v$ represent the p value for the test

Hypothesis to test

We want to determine if the true mean is equal to 200, the system of hypothesis are :

Null hypothesis: $\mu = 200$

Alternative hypothesis: $\mu = 200$

The statistic for this case is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

The statistic is given by:

$t=\frac{201.77-200}{\frac{2.41}{\sqrt{10}}}=2.32$

The degrees of freedom are given by:

$df=n-1=10-1=9$

The p value for this case is given by:

$p_v =2*P(t_{(9)}>2.32)=0.0455$

For this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly different from 200 F.

4 0

2 years ago

A circle with radius of \greenD{6\,\text{cm}}6cmstart color #1fab54, 6, start text, c, m, end text, end color #1fab54 sits insid

Diano4ka-milaya [45]

Answer: $141 cm^{2}$