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Sunny_sXe [5.5K]
3 years ago
12

Please help me! Have to get it done before 5

Mathematics
2 answers:
qwelly [4]3 years ago
8 0
I’m not sure about the first 2
wlad13 [49]3 years ago
6 0

Answer:

I am not sure about the second one but the first question that ask if the data is linear, the answer is no since even though the change in x-values is constant but the change in y-value is inconsistent.

Hope this helps!

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Mrs. Saddler and her family go out for dinner. Their bill was $55.25. Mrs. Saddler leaves a 18% tip. What is Mrs. Saddler’s tota
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2 years ago
An elevator containing five people can stop at any of seven floors. What is the probability that no two people exit at the same
elena-s [515]

Answer:

Approximately 0.15 (360 / 2401.) (Assume that the choices of the 5 passengers are independent. Also assume that the probability that a passenger chooses a particular floor is the same for all 7 floors.)

Step-by-step explanation:

If there is no requirement that no two passengers exit at the same floor, each of these 5 passenger could choose from any one of the 7 floors. There would be a total of 7 \times 7 \times 7 \times 7 \times 7 = 7^{5} unique ways for these 5\! passengers to exit the elevator.

Assume that no two passengers are allowed to exit at the same floor.

The first passenger could choose from any of the 7 floors.

However, the second passenger would not be able to choose the same floor as the first passenger. Thus, the second passenger would have to choose from only (7 - 1) = 6 floors.

Likewise, the third passenger would have to choose from only (7 - 2) = 5 floors.

Thus, under the requirement that no two passenger could exit at the same floor, there would be only (7 \times 6 \times 5 \times 4 \times 3) unique ways for these two passengers to exit the elevator.

By the assumption that the choices of the passengers are independent and uniform across the 7 floors. Each of these 7^{5} combinations would be equally likely.

Thus, the probability that the chosen combination satisfies the requirements (no two passengers exit at the same floor) would be:

\begin{aligned}\frac{(7 \times 6 \times 5 \times 4 \times 3)}{7^{5}} \approx 0.15\end{aligned}.

5 0
2 years ago
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