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Tasya [4]
3 years ago
7

Triangle EFG is dilated by a scale factor of 3 centered at (0, 1) to create triangle E'F'G'. Which statement is true about the d

ilation?
Best answer gets brainliest
E(0,5) F(1,1) G(-2,1) H(0,1)

segment EH and segment E prime H prime both pass through the center of dilation. The slope of segment EF is the same as the slope of segment E prime H prime. segment E prime G prime will overlap segment EG. segment EH ≅ segment E prime H prime.
Mathematics
1 answer:
lesya [120]3 years ago
4 0

Answer:

segment EH and segment E prime H prime both pass through the center of dilation.  

Step-by-step explanation:

Center of dilation is point (0.1), same as H. Both, E(0,5) and H(0,1) are placed over y-axis, then E' and H' are also located at y-axis.

After dilation, H' is placed at (0,1) because it coincides with the center of dilation

Distance between E and center of dilation is 4 units, then E' should be at 4*3=12 units from the center of dilation and over y-axis. Therefore, E' is placed at (0, 13)

So, segment E'H' goes from (0,1) to (0,13), and pass through the center of dilation, like segment EH.

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For a certain​ candy, 15​% of the pieces are​ yellow, 1010​% are​ red, 2020​% are​ blue, 55​% are​ green, and the rest are brown
MatroZZZ [7]
A) the probability it is brown would be 50%; the probability it is yellow or blue would be 35%; the probability it is not green is 95%; the probability it is striped is 0%.
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Explanation:
A) The probability that it is brown is the percentage of brown we have.  Brown is not listed, so we subtract what we are given from 100%:
100-(15+10+20+5) = 100-(50) = 50%.  The probability that one drawn is yellow or blue would be the two percentages added together:  15+20 = 35%.  The probability that it is not green would be the percentage of green subtracted from 100:  100-5=95%.  Since there are no striped candies listed, the probability is 0%.
B) Since we have an infinite supply of candy, we will treat these as independent events.  All 3 being brown is found by taking the probability that one is brown and multiplying it 3 times:
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The probability that none are yellow is found by raising the probability that the first one is not yellow, 100-15=85%=0.85, to the third power:

0.85^3 = 0.614 = 61.4%.

The probability that at least one is green is computed by subtracting 1-(probability of no green).  We first find the probability that all three are NOT green:
0.95^3 = 0.857375
1-0.857375 = 0.143 = 14.3%.
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