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3 years ago

The isosceles triangle theorem says "If two sides of a triangle are congruent,

Mathematics

1 answer:

BartSMP [9]3 years ago

8 0

Answer:

D. Draw KM so that Mis the midpoint of JL, then prove AJKM * A

LKM using SAS.

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What is 1163853739x948096388

Free_Kalibri [48]

Standard calculator will tell you that this is <span>1.1034455e+18 If that makes sense</span>

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3 years ago

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Can someone help me ASAP

lianna [129]

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D. 1/5^15

Step-by-step explanation:

When an exponent is negative you need to get the reciprocal to make it positive.

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In PE class, Josiah and his friends had to measure their heights. They each

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3 1/2 = 1 yard and 1/2 inch then a yard then 31 divided by 12 is 2.??

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3 years ago

Solve for t.<br> - 9r +5 <17<br> AND 132 + 25 <-1

Deffense [45]

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3 years ago

Reading proficiency: An educator wants to construct a 99.5% confidence interval for the proportion of elementary school children

Shkiper50 [21]

Answer:

A sample size of 345 is needed so that the confidence interval will have a margin of error of 0.07

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 of the interval is given by:

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

In this problem, we have that:

$p = 0.69$

99.5% confidence level

So $\alpha = 0.005$ , z is the value of Z that has a pvalue of $1 - \frac{0.005}{2} = 0.9975$ , so $Z = 2.81$ .

Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.07?

This is n when M = 0.07. So

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

$0.07 = 2.81\sqrt{\frac{0.69*0.31}{n}}$

$0.07\sqrt{n} = 1.2996$

$\sqrt{n} = \frac{1.2996}{0.07}$

$\sqrt{n} = 18.5658$

$(\sqrt{n})^{2} = (18.5658)^{2}$

$n = 345$

A sample size of 345 is needed so that the confidence interval will have a margin of error of 0.07

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3 years ago