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aleksandr82 [10.1K]
3 years ago
10

I need the answer to number 2

Mathematics
1 answer:
gladu [14]3 years ago
8 0
Subtract 10 from both sides :
-3y + 10 - 10 = -14 -10
3y = -24
divide both sides with 3 :
y = -8
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7,060,268,214.what is this number rounded to the nerest hundred million?
kodGreya [7K]
7,060,268,214 rounded to the nearest hundred million is 7,100,000,000.

It's because 60,268,214 is closer to 100,000,000 than 000,000,000 (in terms of 7 million of course).
6 0
3 years ago
Please help! Have had 10 attempts and still haven't got it!
adell [148]

Answer:

28

Step-by-step explanation:

To find the area of the shaded region, you have to start off with the area of the whole triangle. To find that, you multiply the height and length. length=10 height=8 (10x8=80). Then, since it's a triangle, you have to divide it by two (80/2=40). So, 40 is the area of the whole triangle. Then, you find the area of the white region, which is 12 (3x4=12). Finally, you take the area of the white square and subtract it from the area of the whole triangle (80-12=28). You would get 28.

5 0
3 years ago
What is the parallel slope of m= -2
Anna35 [415]
You mean what is the slope of a line parallel to the line whose slope is -2?

The answer is -2... the slopes of parallel lines are equal
3 0
3 years ago
MATHSWATCH
Jobisdone [24]

Answer:19683x

Step-by-step explanation:

Evaluate the exponent

27.93x

27.729x

i am pretty sure sorry if I am wrong

7 0
3 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
3 years ago
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