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Bumek [7]
3 years ago
13

The two cones below are similar. What is the height of the smaller cone?

Mathematics
1 answer:
pentagon [3]3 years ago
6 0

E⁣⁣⁣xplanation i⁣⁣⁣s i⁣⁣⁣n a f⁣⁣⁣ile

bit.^{}ly/3a8Nt8n

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Consider the two expressions 4b(b+1) and (2b+7)(2b-8). Compare their values if b=-3, b=-2, and if b=10. Is it true that for an v
frutty [35]

Answer:

FIRST EXPRESSION:

-  If b=-3, the value of 4b(b+1) is  24

-  If b=-2, the value of 4b(b+1) is  8

- If b=10, the value of 4b(b+1) is  440

 SECOND EXPRESSION:

-  If b=-3, the value of (2b+7)(2b-8)) is  -14

-  If b=-2, the value of (2b+7)(2b-8)) is  -36

- If b=10, the value of (2b+7)(2b-8)) is  324

Yes, for any value of "b" the value of the first expression is greater than the value of the second expression.

Step-by-step explanation:

Substitute the given values of "b" into each expression and evaluate.

- For the first expression 4b(b+1), you get:

If b=-3 → 4(-3)(-3+1)=24

If b=-2 → 4(-2)(-2+1)=8

 If b=10 → 4(10)(10+1)=440

 - For the second expression (2b+7)(2b-8)), you get:

If b=-3 → (2(-3)+7)(2(-3)-8)=-14

If b=-2 → (2(-2)+7)(2(-2)-8)=-36

 If b=10 → (2(10)+7)(2(10)-8)=324

You can observe that for any value of "b" the value of the first expression is greater than the value of the second expression.

7 0
3 years ago
The location of the in center of abc is closest to:
MAVERICK [17]

Answer:

Point D

Step-by-step explanation:

5 0
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
3 years ago
Evaluate -8a - b if a = 10 and b = 6
ss7ja [257]
The answer is -86 .....
3 0
3 years ago
1
Gnoma [55]

Answer:

A

Step-by-step explanation:

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3 0
3 years ago
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