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2 years ago

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A 4" by 6" by 8" rectangular solid is cut by slicing through the midpoint of three adjacent edges. What is the number of inches

in the sum of the lengths of the edges of the tetrahedron that is cut?

1 answer:

Triss [41]2 years ago

7 0

Answer:

22.078

Step-by-step explanation:

I have attached the images above

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Ik this is the easiest question but who wanna do it for me?:(

sladkih [1.3K]

Answer:

B.

Step-by-step explanation:

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2 years ago

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Choose an American Adult at random. The probability that you choose women is 0.25

Pepsi [2]

Answer: This is a statement. This is not a question.

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2 years ago

If a phone card is used to make a long distance phone call, you are charged $0.50 per call plus an additional $0.31 per minute.

Nikitich [7]

Part A: c for cost. c=0.31m+0.5

0.31m is the cost per minute. 0.5 is cost per call.

Part B: 0.31m+0.5=5.15 to solve we must rearrange.

subtract 0.5 from each side giving us 0.31m=4.85

divide by 0.31 giving us m=15.65

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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

2 years ago

6. A publisher requires 2⁄3 of a page of advertisements for every 5 pages in a magazine. If a magazine has 98 pages, to the near

earnstyle [38]

We have (2/3)÷5 = 2/15 advertisments for 1 page;

Then, we have 98 · (2/15) = 12.2(6) advertisments for 98 pages;

The nearest whole page is 12.

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3 years ago