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babunello [35]
2 years ago
10

Please help me with this question!

Mathematics
1 answer:
zalisa [80]2 years ago
3 0

Answer:

The answer would be c

Step-by-step explanation:

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A yogurt costs 45p. How many yogurts can be bought for five pounds?
LekaFEV [45]
1/9 of a yogurt can be bought with 5 pounds.
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Please help! Two math problems!!!
Fynjy0 [20]

Answer:

Tell me if i am wrong. :)

Step-by-step explanation:

To solve for y:

Q.1   2x + y = -1

       y = -1 - 2x

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matrenka [14]

Answer: C. Parabola

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5 0
1 year ago
There are 4 3/8 pounds of bricks in a bag. Each brick weighs 7/8 of a pound. How many bricks are in the bag?
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1 year ago
Read 2 more answers
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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