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3 years ago

Please help with these math questions part one

Mathematics

1 answer:

katrin [286]3 years ago

3 0

V of cone=(1/3)(pi)(h)(r^2)
V of cylinder=(pi)(h)(r^2)

1.
r=6
h=10
v=(1/3)(pi)(10)(6^2)
v=1/3pi10(36)
v=1/3pi360
v=120pi
answer is B

2.
r=d/2
d=10
10/2=5=r
V=(3.14)(5)(5^2)
V=3.14(125)
V=392.5
answer is B

3. 9.2, 3.7
V=(pi)(9.2)(3.7^2)
V=(pi)(9.2)(13.69)
V=(pi)(125.948)
input 125.948

4. v=(1/3)(pi)(12)(3^2)
V=(1/3)(pi)(12)(9)
V=(pi)(12)(3)
V=pi(36
v=36pi
v=36(3.14)
V=113.04

input 113.04

ANSERS
1.B
2.B
3. 129.948
4. 113.04

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P ( B0 / D0 ) = 0.59877

P ( B1 / D0 ) = 0.25793

P ( B2 / D0 ) = 0.14329

Step-by-step explanation:

Given:

- 0 be the event that the batch has 0 defectives = (0 ) = 0.52

- 1 be the event that the batch has 1 defectives = (1 ) = 0.28

- 2 be the event that the batch has 2 defectives = (2 ) = 0.2

- Two components are selected

Find:

What are the probabilities associated with 0, 1, and 2 defective components being in the batch under each of the following conditions?

(a) Neither tested component is defective.

Solution:

Let 0 be the event that neither selected component is defective.

- The event 0 can happen in three different ways:

(i) Our batch of 10 is perfect, and we get no defectives in our sample of two;

P(i) = P(B0) = 0.52

(ii) Our batch of 10 has 1 defective, but our sample of two misses them;

P ( no defect / B1 ) = P ( no defect ) / P ( B 1 )

= 9C2 / 10C2 = 0.8

P ( ii ) = 0.28*0.8 = 0.224

(iii) Our batch has 2 defective, but our sample misses them.

P ( no defect / B2 ) = P ( no defect ) / P ( B 2 )

= 8C2 / 10C2 = 56/90

P ( iii ) = 0.2*56/90 = 0.124444

- Then,

P(Do) = P(i) + P(ii) + P(iii)

P(Do) = 0.52 + 0.224 + 0.124444 = 977/1125

We use the general conditional probability formula:

P ( B0 / D0 ) = P ( B0 & D0 ) / P( D0 )

P ( B0 / D0 ) = 0.52*1125 / 977 = 0.59877

P ( B1 / D0 ) = P ( B1 & D0 ) / P( D0 )

P ( B1 / D0 ) = 0.224*1125 / 977 = 0.25793

P ( B2 / D0 ) = P ( B2 & D0 ) / P( D0 )

P ( B2 / D0 ) = 0.12444*1125 / 977 = 0.14329

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3 years ago