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3 years ago

What is the distance between the points? Round to the nearest tenth if necessary.

Mathematics

1 answer:

Greeley [361]3 years ago

8 0

Answer:

d = 8.602325
Rounded = 9

For:
(X1, Y1) = (1, 4)
(X2, Y2) = (6, -3)

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Answer:

$8 y^{50}$

Step-by-step explanation:

Given (64 y Superscript 100 Baseline) Superscript one-half.

Let us write it into an equation.

$\left(64 y^{100}\right)^{\frac{1}{2}}$

Apply radical rule: $\sqrt[n]{a}=a^{\frac{1}{2}}$ and $a^{m+n}=a^m+a^n$

$\begin{aligned}\left(64 y^{100}\right)^{\frac{1}{2}} &=\sqrt[2]{64 y^{100}} \\&=\sqrt[2]{8^{2} y^{50} y^{50}} \\&=\sqrt[2]{8^{2}\left(y^{50}\right)^{2}} \\&=8 y^{50}\end{aligned}$

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3 years ago

From a large number of actuarial exam scores, a random sample of scores is selected, and it is found that of these are passing s

Mnenie [13.5K]

<u>Supposing 60 out of 100 scores are passing scores</u>, the 95% confidence interval for the proportion of all scores that are passing is (0.5, 0.7).

The lower limit is 0.5.
The upper limit is 0.7.

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

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

In which

z is the z-score that has a p-value of $\frac{1+\alpha}{2}$ .

60 out of 100 scores are passing scores, hence $n = 100, \pi = \frac{60}{100} = 0.6$

95% confidence level

So $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so $z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6 - 1.96\sqrt{\frac{0.6(0.4)}{100}} = 0.5$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6 + 1.96\sqrt{\frac{0.6(0.4)}{100}} = 0.7$

The 95% confidence interval for the proportion of all scores that are passing is (0.5, 0.7).

The lower limit is 0.5.
The upper limit is 0.7.

A similar problem is given at brainly.com/question/16807970

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3 years ago