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Leno4ka [110]
2 years ago
12

A parking lot has three colors of cars parked in it. There are 120 white cars parked in the lot. There are 80 more red cars park

ed in the lot than there are white cars. There are 40 fewer black cars parked in the lot than white cars. What is the ratio of white cars to red cars to black cars? Write the ratio in simplest form.
Mathematics
1 answer:
ElenaW [278]2 years ago
4 0
The ratio should be 3:5:2
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Tom sold 22 tickets and Steve sold 57 tickets here is the method

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Solve this:/ 2m - 5 - 5 = -3m + 15
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3 years ago
Individuals filing federal income tax returns prior to March 31 had an average refund of $1102. Consider the population of "last
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Answer:

a) The null hypothesis states that the last-minute filers average refund is equal to the early filers refund. The alternative hypothesis states that the last-minute filers average refund is less than the early filers refund.

H_0: \mu=1102\\\\H_a:\mu < 1102

b) P-value = 0.0055

c) The null hypothesis is rejected.

There is enough statistical evidence to support the claim that that the refunds of the individuals that wait until the last five days to file their returns is on average lower than the early filers refund.

d) Critical value tc=-1.96.

As t=-2.55, the null hypothesis is rejected.

Step-by-step explanation:

We have to perform a hypothesis test on the mean.

The claim is that the refunds of the individuals that wait until the last five days to file their returns is on average lower than the early filers refund ($1102).

a) The null hypothesis states that the last-minute filers average refund is equal to the early filers refund. The alternative hypothesis states that the last-minute filers average refund is less than the early filers refund.

H_0: \mu=1102\\\\H_a:\mu < 1102

b) The sample has a size n=600, with a sample refund of $1050 and a standard deviation of $500.

We can calculate the z-statistic as:

t=\dfrac{\bar x-\mu}{s/\sqrt{n}}=\dfrac{1050-1102}{500/\sqrt{600}}=\dfrac{-52}{20.41}=-2.55

The degrees of freedom are df=599

df=n-1=600-1=599

The P-value for this test statistic is:

P-value=P(t

c) Using a significance level α=0.05, the P-value is lower than the significance level, so the effect is significant. The null hypothesis is rejected.

There is enough statistical evidence to support the claim that that the refunds of the individuals that wait until the last five days to file their returns is on average lower than the early filers refund.

d) If the significance level is α=0.025, the critical value for the test statistic is  t=-1.96. If the test statistic is below t=-1.96, then the null hypothesis should be rejected.

This is the case, as the test statistic is t=-2.55 and falls in the rejection region.

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The mean of a set of 5 numbers is 39 and the median is 40. If the mean of the largest two numbers is 49 , then what is the sum o
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