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scoray [572]
3 years ago
6

Help!! Lots of points

Mathematics
2 answers:
GuDViN [60]3 years ago
6 0

Answer:

The third option ||| Evaluate 5^2 and 2^2 ||| and the last option ||| Rewrite the expression as (5/2)^2 ||| are the correct choices.

So, C and E

murzikaleks [220]3 years ago
3 0

\frac{ {5}^{2} }{ {2}^{2} }  =  {( \frac{5}{2}) }^{2}
Since they are both squared.

Evaluating 5^2 and 2^2 and finding the quotient also works.

\frac{ {5}^{2} }{ {2}^{2} } =  \frac{5 \times 5}{2 \times 2}   =  \frac{25}{4}  = 6.25\\  \frac{ {5}^{2} }{ {2}^{2} }  =  {( \frac{5}{2} )}^{2}  = 2.5 \times 2.5 = 6.25
The third option, "Evaluate 5^2 and 2^2 ...," and the last option, "Rewrite the expression as (5/2)^2 ...," are the correct choices.
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I’m having a hard time with this. Is this a typo by the teacher or should I actually be looking for a size?
Lemur [1.5K]

Step-by-step explanation:

There are 12 games in the population.  You need to use a random number generator to choose 2 of these games.

RandomSample[{1,2,3,4,5,6,7,8,9,10,11,12},2]

Let's say the first sample you get is {1,5}.  That corresponds to game times of 8 minutes and 7 minutes.  The mean game time for that sample is 7.5 minutes.  So the first row in your table would be:

\left[\begin{array}{ccc}Sample&List\ of\ Game\ Times&Mean\ Game\ Time\\1&8,7&7.5\end{array}\right]

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3 years ago
What is the value of x?
mel-nik [20]

Answer:

x = 24

Step-by-step explanation:

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3 years ago
The probability that your call to a service line is answered in less than 30 seconds is 0.75. Assume that your calls are indepen
vfiekz [6]

Answer:

a) 0.2581

b) 0.4148

c) 17

Step-by-step explanation:

For each call, there are only two possible outcomes. Either they are answered in less than 30 seconds. Or they are not. The probabilities for each call are independent. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

In this problem we have that:

p = 0.75

a. If you call 12 times, what is the probability that exactly 9 of your calls are answered within 30 seconds? Round your answer to four decimal places (e.g. 98.7654).

This is P(X = 9) when n = 12. So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 9) = C_{12,9}.(0.75)^{9}.(0.25)^{3} = 0.2581

b. If you call 20 times, what is the probability that at least 16 calls are answered in less than 30 seconds? Round your answer to four decimal places (e.g. 98.7654).

This is P(X \geq 16) when n = 20

So

P(X \geq 16) = P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 16) = C_{20,16}.(0.75)^{16}.(0.25)^{4} = 0.1897

P(X = 17) = C_{20,17}.(0.75)^{17}.(0.25)^{3} = 0.1339

P(X = 18) = C_{20,18}.(0.75)^{18}.(0.25)^{2} = 0.0669

P(X = 19) = C_{20,19}.(0.75)^{19}.(0.25)^{1} = 0.0211

P(X = 20) = C_{20,20}.(0.75)^{20}.(0.25)^{0} = 0.0032

So

P(X \geq 16) = P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.1897 + 0.1339 + 0.0669 + 0.0211 + 0.0032 = 0.4148

c. If you call 22 times, what is the mean number of calls that are answered in less than 30 seconds? Round your answer to the nearest integer.

The expected value of the binomial distribution is:

E(X) = np

In this question, we have n = 22

So

E(X) = 22*0.75 = 16.5

The closest integer to 16.5 is 17.

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