Need help please!!!!!!

You are driving 2760 miles across the country. During the first 3 days of your trip, you drive 1380 miles. If you continue to dr

prohojiy [21]

The entire trip will take 6 days.

(Encircle this answer, as said on the directions)

Step-by-step explanation:

We know that the first three days of the trip, we traveled 1380 miles.

Find the UNIT RATE of MILES PER day:

1380/3 = 460

Unit rate = 460 miles per day

If we were to drive at this (460 mi per day) rate EACH day and the whole journey takes 2760 miles across the country.

FInd the NUMBER of DAYS of the ENTIRE TRIP:

2760/460 = 6

It will take us 6 days to drive to the destination.

5 0

2 years ago

3,152 divided by 16 is what

juin [17]

The quotient is<em> 197</em> .

You have a choice of 3 tools to use, any one of which will return the same answer:

-- your calculator

-- your pencil

-- your brain

7 0

3 years ago

Read 2 more answers

These are so random? I don’t know why I have to put these? Ignore them

Maslowich

What are so random? You dont know why you have to put what? Lol XD

7 0

2 years ago

In a college student poll, it is of interest to estimate the proportion p of students in favor of changing from a quarter-system

Contact [7]

Answer:

n=206

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ . And the critical value would be given by:

$t_{\alpha/2}=-2.58, t_{1-\alpha/2}=2.58$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.09$ , we can use as prior estimate of p 0.5, since we don't have any other info provided, and we are interested in order to find the value of n, if we solve n from equation (a) we got: