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3 years ago

Find the other endpoint of the line segment with the given endpoint and mid point.

1 answer:

6 0

So base on your question that ask to provide and end point of the four statement you give:
1.Endpoint(-1,9), midpoint:(-9,-10) - (-17, -29)
2.Endpoint(2,5), midpoint(5,1) - (4,5)
3.Endpoint(9,-10), midpoint(4,8) - i cant determine sorry
4.Find the point that is one-fourth of the way from (2,4) to (10,8) - also i cant.

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The first five terms of a pattern are shown.

Andrej [43]

Answer:

Step-by-step explanation:

There is a common ratio between consecutive term in the sequence, that is

$\frac{3}{10}$ ÷ $\frac{3}{100}$ = 3 ÷ $\frac{3}{10}$ = 30 ÷ 3 = 300 ÷ 30 = 10

This indicates the sequence is geometric with n th term

$a_{n}$ = a $(r)^{n-1}$

where a is the first term and r the common ratio

Here a = $\frac{3}{100}$ and r = 10 , thus

$a_{n}$ = $\frac{3}{100}$ $(10)^{n-1}$

= $\frac{3}{10^{2} }$ × $10^{n-1}$

= 3 × $10^{-2}$ × $10^{n-1}$

= 3 × $10^{n-3}$

Thus

$a_{n}$ = 3 $(10)^{n-3}$ → C

6 0

4 years ago

Mrs.Benite earns $3,400 every month.She spends 25% of her drawings on rent and 30% of the remainder on transportation.How much d

lutik1710 [3]

25% of 3,400 is 3,400/4=$850 spent rent
3,400-850= 2,550 remaining
30% of 2,550= 2550•0.3= $765 spent on transportation

4 0

3 years ago

A candidate for a US Representative seat from Indiana hires a polling firm to gauge her percentage of support among voters in he

UkoKoshka [18]

Answer:

a) The minimum sample size is 601.

b) The minimum sample size is 2401.

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:

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

For this problem, we have that:

We dont know the true proportion, so we use $\pi = 0.5$ , which is when we are are going to need the largest sample size.

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 $Z = 1.96$ .

a. If a 95% confidence interval with a margin of error of no more than 0.04 is desired, give a close estimate of the minimum sample size that will guarantee that the desired margin of error is achieved. (Remember to round up any result, if necessary.)

This is n for which M = 0.04. So

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

$0.04 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.04\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.04}$

$(\sqrt{n})^2 = (\frac{1.96*0.5}{0.04})^2$

$n = 600.25$

Rounding up

The minimum sample size is 601.

b. If a 95% confidence interval with a margin of error of no more than 0.02 is desired, give a close estimate of the minimum sample size necessary to achieve the desired margin of error.

Now we want n for which M = 0.02. So

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

$0.02 = 1.96\sqrt{\frac{0.5*0.5}{n}}$