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Irina-Kira [14]
3 years ago
9

The frequency table shows the favorite college football teams of middle school students.What fraction of the students chose soon

ers? write the fraction a decimal.

Mathematics
2 answers:
elixir [45]3 years ago
7 0

Answer:

i) Fraction of the students chose Sooners = \frac{5}{20}

ii)  \frac{5}{20} = 0.25

Step-by-step explanation:

The total number of teams = 5

The total number of students chose the football teams = 3 + 6 + 5+ 2 + 4 = 20

The number of students chose Sooners team = 5

Fraction of the students chose Sooners = \frac{5}{20}

Now we have to convert the fraction \frac{5}{20} to decimal using division.

When we divided 5 by 20, we get 0.25

Therefore, the answer is

i) Fraction of the students chose Sooners = \frac{5}{20}

ii)  \frac{5}{20} = 0.25

masya89 [10]3 years ago
4 0
The fraction would be 5/20=1/4=.25 in decimal form.

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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

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_{4,0}.(0.96)^{0}.(0.04)^{4} = 0.00000256

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.00000256 = 0.99999744

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

7 0
3 years ago
HELP PLEASE!!!!!!!!!!!!
Serggg [28]
C = total cents
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I think that's it. Hope it helps you.
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Answer:

A=54m²

Step-by-step explanation:

A = hbb/2 =9 · 12/2 = 54m²

8 0
3 years ago
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