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Tatiana [17]
4 years ago
15

What is the name of the segment inside the large triangle?

Mathematics
2 answers:
NNADVOKAT [17]4 years ago
6 0
The segment inside the large triangle is B: Altitude.
Arlecino [84]4 years ago
5 0
ANSWER

The name of the segment inside the triangle is
altitude.

EXPLANATION

The segment inside the large triangle is a line from one vertex of the large triangle that meets the the opposite side of the large triangle at right angle.


Any line from one vertex of a triangle that meets the side opposite to that vertex at right angle is referred to as the altitude of the triangle.

Sometimes the altitude may fall outside the triangle.


Therefore the correct answer is option 2.

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61 = 3b + 7

3b + 7 = 61

3b = 61 - 7

3b = 54

b = 54/3

b = 18.
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Circle A has been enlarged to create circleA prime . The picture below shows the diameters of both circles.
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Which choice below is a common multiple of 8 and 10 (but not the least common multiple).
Ilya [14]

Answer:

160

Step-by-step explanation:

4 is too low, 90 cant be divided by 8 and 160 is the only answer even though it is highest number

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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

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3 years ago
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Answer:

$10,000

Step-by-step explanation:

Cross multiply and divide by 55.

Hope this helps!

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