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3 years ago

Help me with Geometry

Mathematics

1 answer:

Sergeeva-Olga [200]3 years ago

5 0

$\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
P&({{ 3}}\quad ,&{{ 5}})\quad 
% (c,d)
Q&({{ c}}\quad ,&{{ d}})
\end{array}\quad
% coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)=(-2,0)\\\\
-----------------------------\\\\
\left(\cfrac{{{ c}} + {{ 3}}}{2}\quad ,\quad \cfrac{{{ d}} + {{ 5}}}{2} \right)=(-2,0)M\implies 
\begin{cases}
\cfrac{c+3}{2}=-2\\\\
\cfrac{d+5}{2}=0
\end{cases}$

if you solve those equations for "c" and "d" respectively, you'd get, well the Q point of c,d

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Step-by-step explanation:

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3 years ago

What is the slope of the line on the graph?

Serga [27]

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Graph is not given but we can still find the answer.

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3 years ago

Read 2 more answers

The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e

DanielleElmas [232]

Answer:

The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.13, 0.168).

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 $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

1291 tenth graders, 1098 read above the eighth grade level.

1291 - 1098 = 193 read at or below this level.

We want the 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level.

So $n = 1291, \pi = \frac{193}{1291} = 0.149$

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 $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.149 - 1.96\sqrt{\frac{0.149*0.851}{1291}} = 0.13$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.149 - 1.96\sqrt{\frac{0.149*0.851}{1291}} = 0.168$

The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.13, 0.168).

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3 years ago