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Llana [10]
3 years ago
5

At the toll booth, the chance that a driver has exact change is 1/4, independent of vehicle. Find the probability that no vehicl

e has exact change in the first 10 minutes.
Mathematics
1 answer:
MAXImum [283]3 years ago
4 0

Answer:

g

Step-by-step explanation:

 b

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Why is #2 the expression ln(2e)^3x equivalent to choice B?
motikmotik
Hello:
ln(2e^3x) =ln2+lne^3x
<span>                =ln2 +3xlne
                 = ln2+3x        .....lne=1       
               </span><span>choice B</span>
8 0
3 years ago
URGENT PLEASE HURRY I'M RUNNING OUT OF TIME!!!!!
tester [92]
2.15
1hr is 60 min so
60÷15=4
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45÷4=11.25
90+11.25=101.25 miles
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7 0
4 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

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
In the end it says solve for X <br> show work
elena55 [62]

Answer:

x = 53

Step-by-step explanation:

angle 3 + angle 8 = 180 because of linear pair

66 + 2x + 8 = 180

74 + 2x = 180

2x = 180 - 74

2x = 106

x = 106/2

x = 53

3 0
2 years ago
Read 2 more answers
Type the correct answer in each box.
Leokris [45]
1 pound= $0.50
10 pounds= $5.00
3 0
3 years ago
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