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2 years ago

Simplify fully 13/2x - 5/x

2 answers:

8 0

Hello There!

Find the LCD:

It is 2x

13/2x - 10/2x

The answer is 3/2x

Hope This Helps You!
Good Luck :)

- Hannah ❤

il63 [147K]2 years ago

4 0

<span>Simplify fully 13/2x - 5/x
We have 13/2x- 5/x= 13/2x- 10/2x= (13-10)/2x= 3/(2x)
Have fun</span>

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A game involves spinning the spinner. What is the probability of the pointer landing on R

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The answer would be A. 1/4.

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A train traveled 30 miles at an average speed of 60 mi/h ind the time taken for the trip

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1/2 hour. 30/60 = 1/2 3 divided by 6 is 1/2 since it's not possible for it to be 1

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What is the estimate of 212,514+396,705

Evgesh-ka [11]

Change 212,514 into 215,000.

Change 396,705 into 400,000.

--------

Estimate:

215,000 + 400,000 = 615,000

Therefore:

212,514 + 396,705 ≈ 615,000

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2 years ago

Find the values of x and y

julsineya [31]

Answer:

y = 70 degrees

x = 11

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The side opposite x + 2 is equivalent so they will both be equal to 13.

To get the meaning of x, simply subtract 2 from 13 giving you x = 11.

Y is across from corner C, which is 70 degrees. These angles are equivalent to each other, meaning that angle y is also 70 degrees.

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2 years ago

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An urn contains 5 white and 10 black balls. A fair die is rolled and that number of balls is randomly chosen from the urn. What

galina1969 [7]

Answer:

Part A:

The probability that all of the balls selected are white:

$P(A)=\frac{1}{6}(\frac{1}{3}+\frac{2}{21}+\frac{2}{91}+\frac{1}{273}+\frac{1}{3003}+0)\\ P(A)=\frac{5}{66}=0.075757576$

Part B:

The conditional probability that the die landed on 3 if all the balls selected are white:

$P(D_3|A)=\frac{\frac{2}{91}*\frac{1}{6}}{\frac{5}{66} } \\P(D_3|A)=\frac{22}{455}=0.0483516$

Step-by-step explanation:

A is the event all balls are white.

D_i is the dice outcome.

Sine the die is fair:

$P(D_i)=\frac{1}{6}$ for i∈{1,2,3,4,5,6}

In case of 10 black and 5 white balls:

$P(A|D_1)=\frac{5_{C}_1}{15_{C}_1} =\frac{5}{15}=\frac{1}{3}$

$P(A|D_2)=\frac{5_{C}_2}{15_{C}_2} =\frac{10}{105}=\frac{2}{21}$

$P(A|D_3)=\frac{5_{C}_3}{15_{C}_3} =\frac{10}{455}=\frac{2}{91}$

$P(A|D_4)=\frac{5_{C}_4}{15_{C}_4} =\frac{5}{1365}=\frac{1}{273}$

$P(A|D_5)=\frac{5_{C}_5}{15_{C}_5} =\frac{1}{3003}=\frac{1}{3003}$

$P(A|D_6)=\frac{5_{C}_6}{15_{C}_6} =0$

Part A:

The probability that all of the balls selected are white:

$P(A)=\sum^6_{i=1} P(A|D_i)P(D_i)$

$P(A)=\frac{1}{6}(\frac{1}{3}+\frac{2}{21}+\frac{2}{91}+\frac{1}{273}+\frac{1}{3003}+0)\\ P(A)=\frac{5}{66}=0.075757576$

Part B:

The conditional probability that the die landed on 3 if all the balls selected are white:

We have to find $P(D_3|A)$

The data required is calculated above:

$P(D_3|A)=\frac{P(A|D_3)P(D_3)}{P(A)}\\ P(D_3|A)=\frac{\frac{2}{91}*\frac{1}{6}}{\frac{5}{66} } \\P(D_3|A)=\frac{22}{455}=0.0483516$

7 0

2 years ago