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Scilla [17]
3 years ago
5

A baseball team played 32 games and lost 8. Katy was the catcher in 5/8 of the winning games and 1/4 of nthe losing games.

Mathematics
1 answer:
kaheart [24]3 years ago
5 0

Answer:

fraction of the games did the team win is = \frac{3}{4}

Katy play game  = 17

Step-by-step explanation:

given data

baseball team played = 32 games

lost = 8

Katy catcher = 5/8 of winning

Katy catcher =  1/4  of losing

to find out

What fraction of the games did the team win and  how many games did Katy play

solution

we know total game = 32 and lost 8

so win is total - lost = 32 - 8 = 24

so

fraction of the games did the team win is = \frac{\frac{24}{8} }{\frac{32}{8} }  

fraction of the games did the team win is = \frac{3}{4}

and

so total game 24 winning is  \frac{5}{8}

so katy winning is  24*\frac{5}{8}

katy winning = 3 × 5 = 15

and

losing  \frac{1}{4}

total lose game 8

so Katy  lose = 8 ×  \frac{1}{4} = 2

so

Katy play game = 15 + 2 = 17

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Answer:

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Step-by-step explanation:

Given expression is  

\frac{2(2+3)}{x(x-1)}\times \frac{4x(x+2)}{10(x+3)}

To find the product of two given fractions as below :

\frac{2(2+3)}{x(x-1)}\times \frac{4x(x+2)}{10(x+3)}=\frac{2(5)}{x(x-1)}\times \frac{2x(x+2)}{5(x+3)}

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=\frac{20x(x+2)}{5x(x-1)(x+3)}

=\frac{20x^2+40x}{(5x^2-5x)(x+3)} (multiply each term in the factor to each term in the another factor )

=\frac{20x^2+40x}{5x^3+15x^2-5x^2-15x}  ( adding the like terms )

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Therefore \frac{2(2+3)}{x(x-1)}\times \frac{4x(x+2)}{10(x+3)}=\frac{4(x+2)}{(x-1)(x+3)}

Therefore the product of given two fractions is \frac{4(x+2)}{(x-1)(x+3)}

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Step-by-step explanation:

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