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zavuch27 [327]
3 years ago
5

HELP ASAP

Mathematics
1 answer:
Keith_Richards [23]3 years ago
7 0
25/6=60/x

x=(60*6)/25
x=14.4sec
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V = .5 x.5 x.5 = .125
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y=x vertical compression compression by a factor of 1/7 please compress the equation of y=x by a vertical compression of 1/7
damaskus [11]

Answer: y = 1/7(x) or y = x/7

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6 0
3 years ago
39​% of college students say they use credit cards because of the rewards program. you randomly select 10 college students and a
cluponka [151]
<span>We use the binomial distribution, which states that:
Probability(r out of n) = (nCr) (p)^r (q)^(n-r)
In this case, n = 10 students, p = 39% = 0.39, and q = 1 - 0.39 = 0.61
a) For r = 2: Probability(2/10) = (10C2) (0.39)^2 (0.61)^(10-2) = 45(0.39)^2 (0.61)^8 = 0.1312, which is the probability for exactly 2.
b) We can first find the probability for r = 0 and r = 1, then subtract that from 1 to determine the probability of at least 2.
For r = 0: </span><span>Probability(0/10) = (10C0) (0.39)^0 (0.61)^(10-0) = (1)(1)(0.61)^10 = 0.0071
</span><span>For r = 1: Probability(1/10) = (10C1) (0.39)^1 (0.61)^(10-1) = (10)(0.39)(0.61)^9 = 0.0456</span>
Then P(0/10) + P(1/10) = 0.0527, so P(at least 2/10) = 1 - 0.0527 = 0.9473.
(c) We have P(2/10) = 0.1312, and we can calculate for the rest similarly:
For r = 3: <span>Probability(3/10) = (10C3) (0.39)^3 (0.61)^(10-3) = 0.2237
</span><span>For r = 4: <span>Probability(4/10) = (10C4) (0.39)^4 (0.61)^(10-4) = 0.2503
</span></span><span>For r = 5: <span>Probability(5/10) = (10C5) (0.39)^5 (0.61)^(10-5) = 0.1920
Therefore the sum of P(2) up to P(5) is 0.7972, so this is the probability of having between 2 to 5 inclusive.
</span></span>
3 0
3 years ago
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