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3 years ago

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3/8 divided by 4????

1 answer:

Alexxx [7]3 years ago

4 0

Answer:

um Yess?

Step-by-step explanation:

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Expand (x^2-3/2)^9 using binomial process

BaLLatris [955]

Use the Binomial expansion theorem to find and simplify each term.

$x^{18} - \frac{ 27x^{6} }{2}+ 81x^{14}- \frac{ 567x^{12} }{2}+ \frac{ 5103x^{10} }{16}- \frac{ 15309x^{8} }{16}+ \frac{ 15309x^{6} }{16} - \frac{ 59049x^{2} }{256}-$ $\frac{19683}{512}$

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3 years ago

Find the measurement of angel ABE

Lostsunrise [7]

Answer is 129(A) hope this helps:)

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2 years ago

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

Find the solution to the system of equations x + y = 9 and x - y = -1.

Galina-37 [17]

Answer:

(4, 5 )

Step-by-step explanation:

x + y = 9 → (1)

x - y = - 1 → (2)

adding the 2 equations term by term will eliminate y

2x + 0 = 8

2x = 8 ( divide both sides by 2 )

x = 4

substitute x = 4 into either of the 2 equations and solve for y

substituting into (1)

4 + y = 9 ( subtract 4 from both sides )

y = 5

solution is (4, 5 )

6 0

1 year ago

Read 2 more answers

Hahaha I need help again please

attashe74 [19]

It could be 2+1 beacuse it says there are twice as many

5 0

3 years ago