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kipiarov [429]
3 years ago
11

Ella grew 3 over 4 of an inch last year. Ben grew 2 over 3 of an inch last year. How much more did Ella grow than Ben last year?

Mathematics
1 answer:
IrinaK [193]3 years ago
7 0

Answer:

Ella grew \frac{1}{12} of an inch more

Step-by-step explanation:

For this problem, you want to subtract Ben's growth from Ella's.

This expression would be \frac{3}{4}-\frac{2}{3}.

To work these out, they must have the same denominator (which should be the lowest common multiple or LCM).

In this case, the LCM is 12, so you want to multiply each side so the denominator is 12. This means the first fraction should be multiplied by \frac{3}{3} and the second by \frac{4}{4}.

This makes the expression \frac{9}{12}-\frac{8}{12}, which equals \frac{1}{12}.

**This content involves adding and subtracting fractions, which you may want to revise. I'm always happy to help!

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There is a 2.7% chance that  a particular death is due to an automobile accident?

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3 years ago
I cant find out the answer for<br> the life of me
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Answer:

  • A. h = (S/2πr) - r

Step-by-step explanation:

<u>Solve the given for h:</u>

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A computer assembling company receives 24% of parts from supplier X, 36% of partsfrom supplier Y, and the remaining 40% of parts
Ivan

Answer:

thus the probability that a part was received from supplier Z , given that is defective is 5/6 (83.33%)

Step-by-step explanation:

denoting A= a piece is defective , Bi = a piece is defective from the i-th supplier and Ci= choosing a piece from the the i-th supplier

then

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P(A)= ∑ 5/100 * 24/100 + 10/100 * 36/100 + 6/100 * 40/100 = 9/125

from the theorem of Bayes

P(Cz/A)= P(Cz∩A)/P(A)

where

P(Cz/A) = probability of choosing a piece from Z , given that a defective part was obtained

P(Cz∩A)= probability of choosing a piece from Z that is defective = P(Bz) = 6/100

therefore

P(Cz/A)= P(Cz∩A)/P(A) = P(Bz)/P(A)= 6/100/(9/125) = 5/6 (83.33%)

thus the probability that a part was received from supplier Z , given that is defective is 5/6 (83.33%)

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