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3 years ago

High Jump: If a person high jumps 6 feet 5 inches, how many inches is the jump?

2 answers:

7 0

If a person jumps 6 feet 5 inches than they are jumping 77 inches. You find this because there are 12 inches in a foot, so you multiply 12 times 6 and get 72. You than add 72 plus 5 and get 77.

Your answer is: 77 inches

Have an amazing day!

Dmitrij [34]3 years ago

4 0

Convert the 6 feet into inches and then add the 5 inches.

1 feet = 12 inches

$\frac{6 feet}{1} * \frac{12 inches}{1 feet}$

$\frac{72 inches}{1}$

Then add the 5 inches to the 72 inches.

5 + 72 = 77

So, 77 inches is the answer.

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LekaFEV [45]

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3 years ago

Congress regulates corporate fuel economy and sets an annual gas mileage for cars. A company with a large fleet of cars hopes to

attashe74 [19]

Answer:

a) Null hypothesis: $\mu \leq 30.2$

Alternative hypothesis: $\mu > 30.2$

b) $X \sim N(\mu=32.12, \sigma=4.83)$

And the distribution for the random sample is given by:

$\bar X \sim N(\mu=32.12, \frac{\sigma}{\sqrt{n}}=\frac{4.83}{\sqrt{50}}=0.683)$

c) $t=\frac{32.12-30.2}{\frac{4.83}{\sqrt{50}}}=2.81$

$p_v =P(t_{(49)}>2.81)=0.0035$

d) If we compare the p value and the significance level assumed $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is significantly higher than 30.2.

e) The p-value is the probability of obtaining the observed results of a test, assuming that the null hypothesis is correct. And for this case is a value to accept or reject the null hypothesis.

Step-by-step explanation:

1) Data given and notation

$\bar X=32.12$ represent the sample mean

$s=4.83$ represent the sample standard deviation

$n=50$ sample size

$\mu_o =30.2$ represent the value that we want to test

$\alpha$ 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)

a. Define the parameter and state the hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 30.2 mpg, the system of hypothesis would be:

Null hypothesis: $\mu \leq 30.2$

Alternative hypothesis: $\mu > 30.2$

If we analyze the size for the sample is > 30 but 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".

b. Define the sampling distribution (mean and standard deviation).

Let X the random variable who represent the variable of interest. And we know that the distribution for X is:

$X \sim N(\mu=32.12, \sigma=4.83)$

And the distribution for the random sample is given by:

$\bar X \sim N(\mu=32.12, \frac{\sigma}{\sqrt{n}}=\frac{4.83}{\sqrt{50}}=0.683)$

c. Perform the test and calculate P-value

Calculate the statistic

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

$t=\frac{32.12-30.2}{\frac{4.83}{\sqrt{50}}}=2.81$

P-value

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

$df=n-1=50-1=49$

Since is a one right tailed test the p value would be:

$p_v =P(t_{(49)}>2.81)=0.0035$

d. State your conclusion.

If we compare the p value and the significance level assumed $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is significantly higher than 30.2.

e. Explain what the p-value means in this context.

The p-value is the probability of obtaining the observed results of a test, assuming that the null hypothesis is correct. And for this case is a value to accept or reject the null hypothesis.

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3 years ago

Convert 6 4/5 to an improper fraction.

Marina86 [1]

6 4/5 as improper fraction is 34/5

5 3/4 as improper fraction is 23/4

3 1/3 by 25/4 is 8/15

14/3 by 3 1/2 is 4/3

3 1/4 by 5 1/2 is 13/22

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2 years ago

Help WILL GIVE BRAINLIEST!! At the beginning of year 1, Carlos invested $600 at an annual compound interest rate of 4%. He makes

OleMash [197]

I got D but I use a diffrnt method