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2 years ago

5

Jane has earned a 91, 85, and 84 on her first three quizzes of the semester. If she hopes to have an A quiz average (90 or above

), what is the lowest score Jane can make on her fourth and final quiz?
She cannot earn an A quiz average*****

100

97

95

1 answer:

fenix001 [56]2 years ago

5 0

Answer:

100

Step-by-step explanation:

91+85+84+100 is 360

Divide that by 360, and you get 90, exactly.

I hope this helps!

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Describe how you would estimate the square root of 100

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I am not completely sure but Khan Academy might show how to do that. Or it might at least give an example.

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Answer:

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Step-by-step explanation:

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3 years ago

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The computer that controls a bank's automatic teller machine crashes a mean of 0.6 times per day. What is the probability that,

professor190 [17]

Answer:

0.2103 = 21.03% probability that, in any seven-day week, the computer will crash less than 3 times.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

Mean of 0.6 times a day

7 day week, so $\mu = 7*0.6 = 4.2$

What is the probability that, in any seven-day week, the computer will crash less than 3 times? Round your answer to four decimal places.

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

In which

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-4.2}*(4.2)^{0}}{(0)!} = 0.0150$

$P(X = 1) = \frac{e^{-4.2}*(4.2)^{1}}{(1)!} = 0.0630$

$P(X = 2) = \frac{e^{-4.2}*(4.2)^{2}}{(2)!} = 0.1323$

So

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0150 + 0.0630 + 0.1323 = 0.2103$

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ElenaW [278]

Answer:

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Step-by-step explanation:

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