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musickatia [10]
3 years ago
14

Triangle EFG is dilated by a scale factor of 2/3 to form triangle E'F'G'. What is the measure of side FG?

Mathematics
1 answer:
Illusion [34]3 years ago
4 0

Answer:

Step-by-step explanation:

Since, the given question is incomplete but to solve these kind of questions,

Use the formula of the scale factor,

Scale factor = \frac{\text{Length of one dimension of the dilated image}}{\text{Length of one dimension of the preimage}}

From the question ΔE'F'G' is the image formed by the dilation of preimage ΔEFG.

Therefore, \frac{2}{3}=\frac{F'G'}{FG}

FG = \frac{3}{2}(F'G')

By substituting any value of F'G' given in the question we can evaluate the measure of side FG.

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5 0
3 years ago
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

P(A)= ∑ P(Bi)*P(C) with i from 1 to 3

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%)

8 0
3 years ago
Read 2 more answers
ASAP HELP 10 POINTS YOU WILL GET
Sholpan [36]

Answer:

I'm pretty sure your answer would be #4

Step-by-step explanation:

Hope this Help you

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lbvjy [14]
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If f(x)=x - 1 and g(x) = 5x – 2, then (f.g) (x)=
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Answer:

Step-by-step explanation:

Hello,

If you mean

(f.g)(x)=f(x).g(x)=(x-1)(5x-2)=5x^2-7x+2

If you mean

(fog)(x)=f(g(x))=(5x-2)-1=5x-3

Thanks

7 0
4 years ago
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