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3 years ago

The height of a person and the person's age, is the relationship linear?

1 answer:

8 0

Liniear means that they go up at a constant rate
example, every year you grow 2 inches and you started at 8 inches tall
that would be linear
but that is not the case
generally, the younger you are, the faster you grow
as you age, you don't grow as much anymore :(

so no, it is not linear

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700 miles the train takes 5 hours to a travel the distance at what unit rate is the train traveling?

dalvyx [7]

Answer:

The rate is 140 miles per hour

Step-by-step explanation:

We can use the formula distance is equal to rate times time

The distance is 700 miles and the time is 5 hours

D =rt

700 = r*5

Divide each side by 5

700/5 = 5r/5

140 =r

The rate is 140 miles per hour

5 0

3 years ago

Find m∠U.<br> Write your answer as an integer or as a decimal rounded to the nearest tenth.

maksim [4K]

Answer:

The answer is 66.4

Step-by-step explanation:

hope this helps mark me brainliest if it right please

7 0

3 years ago

My last question on the pre test someone help lol

Annette [7]

Answer:

Step-by-step explanation:

i just know it

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3 years ago

Assume that the president’s job approval is at 43%. Assume the cost of conducting a poll is $2.50 per person. Imagine if you wer

rodikova [14]

Answer:

The cost difference would be of $22,067.5

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

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

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

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

In this question, we have that:

$\pi = 0.43$

I think there was a small typing mistake, and the confidence level was omitted. I will use a 95% confidence level.

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

How many people are needed for a margin of error of 4 percentage points?

This is n when M = 0.04. So

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

$0.04 = 1.96\sqrt{\frac{0.43*0.57}{n}}$

$0.04\sqrt{n} = 1.96\sqrt{0.43*0.57}$

$\sqrt{n} = \frac{1.96\sqrt{0.43*0.57}}{0.04}$

$(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.43*0.57}}{0.04})^{2}$

$n = 588.5$

Rounding up

For a margin of error of 4 percentage points, 589 people will be sampled.

How many people are needed for a margin of error of 1 percentage point?

This is n when M = 0.01. So

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

$0.01 = 1.96\sqrt{\frac{0.43*0.57}{n}}$

$0.01\sqrt{n} = 1.96\sqrt{0.43*0.57}$

$\sqrt{n} = \frac{1.96\sqrt{0.43*0.57}}{0.01}$

$(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.43*0.57}}{0.01})^{2}$

$n = 9415.7$

Rounding up

For a margin of error of 1 percentage point, 9416 people will be sampled.

What would be the cost difference for a poll designed to have a margin of error of 1 percentage point vs. that of a poll designed to have a margin of error of 4 percentage points?

Cost per person $2.50.

Margin of error of 0.01: Cost of 9416*2.50 = $23,540

Margin of error of 0.04: Cost of 589*2.50 = $1472.5

Difference:

23540 - 1472.5 = $22,067.5

The cost difference would be of $22,067.5

7 0

3 years ago

Please help can you solve it

NikAS [45]

1.30 2.700 3.7 4.9,000 5.30,000 6.500 7.800,000 8.6,000

would you like me to answer anymore???

6 0

4 years ago