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kicyunya [14]
3 years ago
12

What's the midpoint of (0,9) and (10,5)

Mathematics
2 answers:
pychu [463]3 years ago
7 0

Step-by-step explanation:

Hey there!

Given points are: (0,9) and (10,5)

Then;

Using midpoint formula;

(x,y) = ( \frac{x1  + x2}{2} , \frac{y1 + y2}{2} )

Put all values.

(x,y) = ( \frac{0 + 10}{2} , \frac{9 + 5}{2} )

Simplify it.

(x,y) = (5,7)

<u>There</u><u>fore</u><u>,</u><u> </u><u>the</u><u> </u><u>midpo</u><u>int</u><u> </u><u>is </u><u>(</u><u>5</u><u>,</u><u>7</u><u>)</u>

<em><u>Hop</u></em><em><u>e</u></em><em><u> it</u></em><em><u> helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

Masja [62]3 years ago
4 0

Answer:

(5,7)

General Formulas and Concepts:

<u>Pre-Algebra</u>

  • Order of Operations: BPEMDAS

<u>Algebra I</u>

  • Midpoint Formula: (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})

Step-by-step explanation:

<u>Step 1: Define</u>

Point (0, 9)

Point (10, 5)

<u>Step 2: Find midpoint</u>

  1. Substitute:                    (\frac{0+10}{2},\frac{9+5}{2})
  2. Add:                              (\frac{10}{2},\frac{14}{2})
  3. Divide:                          (5,7)
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URGENT: statistics: A researcher wishes to estimate the percentage of adults who support abolishing the penny what size sample s
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Using the margin of error for the z-distribution, the sample sizes are given as follows:

a) 822.

b) 1068.

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions is given by:

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

The margin of error is given by:

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

In which:

  • \pi is the sample proportion.
  • z is the critical value.
  • n is the sample size.

We have a 95% confidence level, hence\alpha = 0.95, z is the value of Z that has a p-value of \frac{1+0.95}{2} = 0.975, so the critical value is z = 1.96.

Item a:

The estimate is of \pi = 0.26, hence we solve for n when M = 0.03.

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

0.03 = 1.96\sqrt{\frac{0.26(0.74)}{n}}

0.03\sqrt{n} = 1.96\sqrt{0.26(0.74)}

\sqrt{n} = \frac{1.96\sqrt{0.26(0.74)}}{0.03}

(\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.26(0.74)}}{0.03}\right)^2

n = 821.2

A sample of 822 is needed.

Item b:

No prior estimate, hence \pi = 0.5.

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

0.03 = 1.96\sqrt{\frac{0.5(0.5)}{n}}

0.03\sqrt{n} = 1.96\sqrt{0.5(0.5)}

\sqrt{n} = \frac{1.96\sqrt{0.5(0.5)}}{0.03}

(\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.5(0.5)}}{0.03}\right)^2

n = 1067.11

A sample of 1068 is needed.

More can be learned about the z-distribution at brainly.com/question/25890103

#SPJ1

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