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3 years ago

What is 2/5 divided by 5/7 in fraction form, not decimal

1 answer:

6 0

Ok. In order to do this set up 2/5 divided by 5/7. The way you divide with fractions is by multiplying them but flipping the second fraction into 7/5 <---this is known as the reciprocal. In other words do:

2/5 x 7/5 =
14/25
There you go.

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What is 30% written as a fraction in simplest form? (Lol I asked the wrong question earlier this is what I meant )

igor_vitrenko [27]

Answer:

3/10

Step-by-step explanation:

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4 years ago

Read 2 more answers

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.

6 0

3 years ago

Sven can type 30 words per minute.How many words would he type in 20 minutes?

saveliy_v [14]

30 x 20=600
your answer is that he would type 600 words in 20 minutes.

8 0

3 years ago

PLZZzzzzz Help me ASAP!!!

ziro4ka [17]

Answer:

The estimate is 217.5.

8 0