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kolbaska11 [484]
2 years ago
7

4.8 yd

Mathematics
1 answer:
Anna007 [38]2 years ago
6 0

Answer:

A

Step-by-step explanation:

The composite solid is made up of a cone and a rectangular prism.

Volume of the composite solid = volume of the cone + volume of the rectangular prism

✔️Volume of Cone = ⅓*π*r²*h

Where,

r = 4.8 yd

h = √(6² - 4.8²) = √12.96 = 3.6 yd

Substitute

Volume of cone = ⅓*π*4.8²*3.6

= 86.86 yd²

✔️Volume of rectangular prism = l*b*h

Where,

l = 7 yd

w = 5 yd

h = 4.5 yd

Substitute

Volume of prism = 7*5*4.5 = 157.5 yd²

✔️Volume of composite solid = 86.86 + 157.6 = 244.4 yd² (which is close to 244.36 yd²)

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at x= -2, y=(-2)^2=4, so one point is (-2,4)

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the line crosses the parabola at (-2,4) and (3,9)

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Best of three In a best out of three series played between teams A and B, the team that gets two wins first wins the entire seri
konstantin123 [22]

Answer:

a) P (x = 2 games )  =  0.49 + 0.09 = 0.58

    P ( x = 3 games ) = 0.063 + 0.147 + 0.147 + 0.063 = 0.42

b) = 2.42 ≈ 2 games

c) P (x = 2 games )  =  0.49 + 0.09 = 0.58

Step-by-step explanation:

Team A chance of winning a game in the series. P( team A ) = 70% = 0.7

P ( team B ) = 0.3

probability of series ending after two games = 58% = 0.58

<u>A) Determine the probability distribution of X number of games played in the series</u>

First we have to consider the possible combinations that will decide the series and they are

( A,A ) , ( B,B) , ( A,B,B) , ( A,B,A ) , ( B,A,A ), ( B,A,B)  = 6 Combinations

( A,A ) = 0.7 * 0.7 = 0.49

( B,B ) = 0.3 * 0.3 = 0.09

( A,B,B ) = 0.7 * 0.3 *0.3 = 0.063

( A,B,A ) = 0.147

( B,A,A ) = 0.147

( B,A,B ) = 0.063

The distribution of the games in the series can be either game or three games before the end of the series

P (x = 2 games )  =  0.49 + 0.09 = 0.58

P ( x = 3 games ) = 0.063 + 0.147 + 0.147 + 0.063 = 0.42

<u>B) the expected number of games to be played </u>

∑ x(Px) = ( 2 * 0.58 ) + 3 ( 0.42 ) = 2.42 ≈ 2 games

<u>C)  Verify that the probability that series ends after two games = 58%</u>

sample space of all possible sequences of wins and losses

( A,A ) , ( B,B) , ( A,B,B) , ( A,B,A ) , ( B,A,A ), ( B,A,B)  = 6 Combinations

( A,A ) = 0.7 * 0.7 = 0.49

( B,B ) = 0.3 * 0.3 = 0.09

( A,B,B ) = 0.7 * 0.3 *0.3 = 0.063

( A,B,A ) = 0.147

( B,A,A ) = 0.147

( B,A,B ) = 0.063

hence :

P (x = 2 games )  =  0.49 + 0.09 = 0.58

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