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2 years ago

15

jana ran the first 3.5 miles of a 5 mile race in 1/3 hour what was the rate in miles per hour for this first part of the race? e

xplain how you solved the problem

1 answer:

Grace [21]2 years ago

4 0

2.8 1/3 of 3.5 is .7 .7 - 3.5=2.8

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A survey of middle school students asked if they play video games. The responses are

kow [346]

Answer:

A. 35.69%

B. 62.77%

Step-by-step explanation:

In this case, what we must do is calculate the probability supported with the data in the table.

Part A.

What is the probability that a randomly selected student is a 7th grader?

Here the total is all the students that would be the sum of all the grades, which are 325 students.

the number of 7th graders is 116, so the probability is:

116/325 = 0.3569

that is, 35.69%

Part B: What percentage of the students interviewed play Video games?

The total number of students is the same 325 and of those 204 play video games, therefore:

204/325 = 0.6277

that is, 62.77%

6 0

3 years ago

Sarah claims that the thickness of the spearmint gum she produces is 7.5 one-hundredths of an inch. A quality control specialist

satela [25.4K]

Answer:

$t=\frac{7.55-7.5}{\frac{0.103}{\sqrt{10}}}=1.539$

$p_v =2*P(t_{(9)}>1.539)=0.158$

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis.

Step-by-step explanation:

Data given and notation

Data: 7.65, 7.60, 7.65, 7.7, 7.55, 7.55, 7.4, 7.4, 7.5, 7.5

We can begin calculating the sample mean given by:

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

And the sample deviation given by:

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

And we got:

$\bar X=7.55$ represent the sample mean

$s=0.103$ represent the sample standard deviation

$n=10$ sample size

$\mu_o =7.5$ represent the value that we want to test

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

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is equal to 7.5 or no, the system of hypothesis would be:

Null hypothesis: $\mu = 7.5$

Alternative hypothesis: $\mu \neq 7.5$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{7.55-7.5}{\frac{0.103}{\sqrt{10}}}=1.539$

P-value

The first step is calculate the degrees of freedom, on this case:

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

Since is a two sided test the p value would be:

$p_v =2*P(t_{(9)}>1.539)=0.158$

Conclusion

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis.

3 0

3 years ago

In this combined function, the domain is all real numbers for x except?( x-2) / (x+5)

Sidana [21]

-5 because the denominator can’t equal zero

4 0

2 years ago

A random sample of 864 births in a state included 426 boys. Construct a 95% confidence interval estimate of the proportion of bo

oksian1 [2.3K]

Using the z-distribution, it is found that the 95% confidence interval is (0.46, 0.526), and it does not provide strong evidence against that belief.

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions is given by:

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which:

$\pi$ is the sample proportion.

z is the critical value.

n is the sample size.

In this problem, we have a 95% confidence level, hence $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so the critical value is z = 1.96.

We have that a random sample of 864 births in a state included 426 boys, hence the parameters are given by:

$n = 864, \pi = \frac{426}{864} = 0.493$

Then, the bounds of the interval are given by:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.493 - 1.96\sqrt{\frac{0.493(0.507)}{864}} = 0.46$

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.493 + 1.96\sqrt{\frac{0.493(0.507)}{864}} = 0.526$

The 95% confidence interval estimate of the proportion of boys in all births is (0.46, 0.526). Since the interval contains 0.506, it does not provide strong evidence against that belief.

More can be learned about the z-distribution at brainly.com/question/25890103

4 0

2 years ago

What is 30% of a $24.50 item

Zarrin [17]

$24.50*0.3 = $7.35
The answer is $7.35

6 0

2 years ago

Read 2 more answers