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3 years ago

If YU= YV ST=16 mQS=34 and mRT=98 find each measure will mark brainliest

Mathematics

1 answer:

aksik [14]3 years ago

8 0

Answer:

QR=8

QU=76

Measure of arc ST=114°

Measure of arc QR=114°

Measure of arc XT=57°

Step-by-step explanation:

Given

YU=YV

ST=16

mQS=34

mRT=98

Using Pythagoras theorem,

QR=8

QU=76

QR=ST

So they both span the same arc.

Let the arc=x

360 - QS - RT = QR + ST

Recall QR=ST

360 - 34 - 98 = x + x

228 = 2x

X=228/2

X=114°

Therefore, measure of arc ST and arc QR=114° each

Measure of the span of arc XT = (1/2) of (arc ST)

= (1/2)(114°)

= 57°

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Answer:

The answer is

Step-by-step explanation:

Equation of a line is y = mx + c

where

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To find the equation of the parallel line we must first find the slope of the original line

From the question the original line is

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Comparing with the general equation above

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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:

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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.

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

$(\sqrt{n})^2 = (\frac{1.96*0.5}{0.02})^2$

$n = 2401$

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3 years ago

Can someone please help me on this one question?

Katyanochek1 [597]

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2b = 30
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5(15) - 5c = 60
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10c + d = 60
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6e = 60
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