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3 years ago

Help? 10 Points

Mathematics

2 answers:

Vera_Pavlovna [14]3 years ago

7 0

5) The answer is D
6) The answer is B

Luda [366]3 years ago

3 0

I think the answer is the third one

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Solve both by substitution method (algebra 2) and show work please<br><br> -2x-7y=22<br>-7x-5y=-1

Alex17521 [72]

$\bf \begin{cases}
-2x-7y=22\\
-7x-5y=-1
\end{cases}\qquad \textit{let's solve the \underline{second one} for "x"}
\\\\\\
-7x-5y=-1\implies -7x=5y-1\implies x=\cfrac{5y-1}{-7}
\\\\\\
\boxed{x=\cfrac{1-5y}{7}}\impliedby \textit{now, let's substitute it in the \underline{first one}}
\\\\\\$

$\bf -2\left( \frac{1-5y}{7} \right)-7y=22\implies \cfrac{10y-2}{7}-7y=22
\\\\\\
\textit{now, let's multiply both sides by 7, to toss the denominator}
\\\\\\
7\left( \cfrac{10y-2}{7}-7y \right)=7(22)\implies 10y-2-49y=154
\\\\\\
-39y=156\implies y=\cfrac{154}{-39}\implies \boxed{y=-4}
\\\\\\
\textit{now, let's use that in the \underline{first one}}
\\\\\\
-2x-7(-4)=22\implies -2x+28=22\implies 6=2x\implies \cfrac{6}{2}=x
\\\\\\
\boxed{3=x}$

4 0

2 years ago

Triangles H J K and L M N are shown. The triangles have identical side lengths and angle measures. Triangle H J K is slightly lo

Hitman42 [59]

Answer:

The answer is b : translate K to N and reflect across the line containing JK

Step-by-step explanation:

I just took the quiz

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

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

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