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Aleonysh [2.5K]
2 years ago
8

Y=3r+2 -3x+y=9 How many solutions?

Mathematics
1 answer:
Alecsey [184]2 years ago
7 0

Answer:

The answer is C or the third one

Step-by-step explanation:

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What is the expected frequency of east campus and passed?
ser-zykov [4K]

The expected frequency of east campus and passed is C. 42 students

<h3>How to calculate the value?</h3>

The table for expected frequency is ,

East Campus West Campus Total

Passed (84*100)/22=42 (84*100)/200 =42 84

Failed (116*100)/200=58 (116*100)/22=58 116

Total 100 100 200

Passed = 84×100/200

= 42

Therefore, the expected frequency of East Campus and Passed is 42 students.

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2 years ago
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3 years ago
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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%.
B) the probability of all brown would be 12.5%; the probability that the third one is the first red one drawn is 8.1%; the probability that none are yellow is 61.4%; the probability that at least one is green is 14.3%.

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:
0.5*0.5*0.5 = 0.125 = 12.5%.  
To find the probability that the first one that is red is the third one drawn, we take the probability that it is NOT red, 100-10 = 90% = 0.9, for the first two, and the probability that it IS red, 10% = 0.1, for the last:
0.9*0.9*0.1 = 0.081 = 8.1%.

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%.
3 0
3 years ago
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