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denis23 [38]
3 years ago
12

Line segment GH has endpoints at G(4,-8) and (-2,6). What is the midpoint of the line segment GH

Mathematics
2 answers:
Irina18 [472]3 years ago
4 0

Answer:

(1,-1)

Step-by-step explanation:

Find the middle between the two points:

4 to -2=6, so the middle would be 3

4-3=1

-8 to 6=14, so the middle would be 7

-8+7=-1

Sophie [7]3 years ago
4 0

Answer:

\displaystyle (1,-1)

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Algebra I</u>

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

Step-by-step explanation:

<u>Step 1: Define</u>

Point G(4, -8)

Point H(-2, 6)

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

Simply plug in your coordinates into the midpoint formula to find midpoint

  1. Substitute in points [Midpoint Formula]:                                                       \displaystyle (\frac{4-2}{2},\frac{-8+6}{2})
  2. [Fraction] Subtract/Add:                                                                                 \displaystyle (\frac{2}{2},\frac{-2}{2})
  3. [Fraction] Divide:                                                                                             \displaystyle (1,-1)
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Find the probability that the person is frequently or occasionally involved in charity work.
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Part A:

If a person is chosen at random, the probability that the person is frequently or occassinally involved in charity work is given by

P(being \ frequently \ involved \ or \ being \ occassionally \ involved)\\ \\= \frac{432}{2881} + \frac{904}{2881} = \frac{1336}{2881}=\bold{0.464}



Part B:

If a person is chosen at random, the probability that the person is female or not involved in charity work at all is given by

P(being&#10; \ female \ or \ not \ being \ involved)\\ \\= &#10;\frac{1402}{2881} + \frac{1545}{2881}-\frac{747}{2881} = &#10;\frac{2200}{2881}=\bold{0.764}



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If a person is chosen at random, the probability that the person is female or not frequently involved in charity work is given by

P(being&#10; \ female \ or \ not \ being \ frequently \ involved)\\ \\= &#10;\frac{1402}{2881} + \frac{904}{2881} + \frac{1545}{2881}-\frac{450}{2881}-\frac{747}{2881} = &#10;\frac{2654}{2881}=\bold{0.921}



Part E:

The events "being female" and "being frequently involved in charity work" are not mutually exclusive because being a female does not prevent a person from being frequently involved in charity work.

Indeed from the table, there are 205 females who are frequently involved in charity work.

Therefore, the answer to the question is "No, because 205 females are frequently involved charity work".
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Answer:

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