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Alecsey [184]
3 years ago
10

PLEASE HELP ASAP ! WILL GIVE BRAINLIEST Quadrilateral OPQR is inscribed inside a circle as shown below. What is the measure of a

ngle O? You must show all work and calculations to receive credit.

Mathematics
1 answer:
finlep [7]3 years ago
6 0

Answer:

The measure of angle P is 43°

Step-by-step explanation:

If a quadrilateral is inscribed inside a circle, the sum of opposite angles are always equal to 180 degrees.

The angle POR is opposite to the angle PQR, and the angle OPQ is opposite to the angle ORQ.

So we have that:

mOPQ + mORQ = 180°

y + 3y + 8 = 180

4y = 172

y = 43°

So the measure of angle P is 43°.

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Mrs. Wester bought five Shawn Mendes CD's to give to friends for Christmas. Each CD cost $12.73. How much did she spend, not inc
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2 years ago
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A bag contains red marbles, white marbles, and blue marbles. Randomly choose two marbles, one at a time, and without replacement
dsp73

Answer:

P(First\ White\ and\ Second\ Blue) = \frac{3}{28}

P(Same) = \frac{67}{210}

Step-by-step explanation:

Given (Omitted from the question)

Red = 7

White = 9

Blue = 5

Solving (a): P(First\ White\ and\ Second\ Blue)

This is calculated using:

P(First\ White\ and\ Second\ Blue) = P(White) * P(Blue)

P(First\ White\ and\ Second\ Blue) = \frac{n(White)}{Total} * \frac{n(Blue)}{Total - 1}

<em>We used Total - 1 because it is a probability without replacement</em>

So, we have:

P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{21 - 1}

P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{20}

P(First\ White\ and\ Second\ Blue) = \frac{9*5}{21*20}

P(First\ White\ and\ Second\ Blue) = \frac{45}{420}

P(First\ White\ and\ Second\ Blue) = \frac{3}{28}

Solving (b) P(Same)

This is calculated as:

P(Same) = P(First\ Blue\ and Second\ Blue)\or\ P(First\ Red\ and Second\ Red)\ or\ P(First\ White\ and Second\ White)

P(Same) = (\frac{n(Blue)}{Total} * \frac{n(Blue)-1}{Total-1})+(\frac{n(Red)}{Total} * \frac{n(Red)-1}{Total-1})+(\frac{n(White)}{Total} * \frac{n(White)-1}{Total-1})

P(Same) = (\frac{5}{21} * \frac{4}{20})+(\frac{7}{21} * \frac{6}{20})+(\frac{9}{21} * \frac{8}{20})

P(Same) = \frac{20}{420}+\frac{42}{420} +\frac{72}{420}

P(Same) = \frac{20+42+72}{420}

P(Same) = \frac{134}{420}

P(Same) = \frac{67}{210}

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