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3 years ago

Fifteen bicyclists entered a race. In how many different orders can the top five finish so that awards can be given?

Mathematics

1 answer:

postnew [5]3 years ago

5 0

Answer:

They can finish the race in 15 x 14 x 13 x 12 x 11 = 360360 order.

Step-by-step explanation:

There are 15 bicyclists in the race. The first position would come from the first bicyclist to the 15th. Meaning the first position can be occupied by 15 individuals. The second position can only be occupied by 14 individuals since one person is already first. Same way the third, fourth and fifth positions can be occupied by 13, 12 and 11 individuals respectively.

Therefore, they can finish the race in

15 x 14 x 13 x 12 x 11 = 360360 order.

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I have been stuck on the problem for so long. please help me:

Xelga [282]

Answer:

D-2

Step-by-step explanation:

You can use elimination to get the value of y. Move numbers to one side and x and ys to the other side. To use elimination, multiply the first equation with 2. You will get 2x+2y=24. add it with the equation below. y will be eliminated and you will get the value of x. X=10. Plug in x and get the y value. Y=2

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3 years ago

Assume that there is a 4% rate of disk drive failure in a year. a. If all your computer data is stored on a hard disk drive wit

kap26 [50]

Answer:

a) 99.84% probability that during a year, you can avoid catastrophe with at least one working drive

b) 99.999744% probability that during a year, you can avoid catastrophe with at least one working drive

Step-by-step explanation:

For each disk drive, there are only two possible outcomes. Either it works, or it does not. The disks 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.

4% rate of disk drive failure in a year.

This means that 96% work correctly, $p = 0.96$

a. If all your computer data is stored on a hard disk drive with a copy stored on a second hard disk drive, what is the probability that during a year, you can avoid catastrophe with at least one working drive?

This is $P(X \ geq 1)$ when $n = 2$

We know that either none of the disks work, or at least one does. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$

We want $P(X \geq 1)$ . So

$P(X \geq 1) = 1 - P(X = 0)$

In which

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

$P(X = 0) = C_{2,0}.(0.96)^{0}.(0.04)^{2} = 0.0016$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0016 = 0.9984$

99.84% probability that during a year, you can avoid catastrophe with at least one working drive

b. If copies of all your computer data are stored on four independent hard disk drives, what is the probability that during a year, you can avoid catastrophe with at least one working drive?

This is $P(X \ geq 1)$ when $n = 4$