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fiasKO [112]
3 years ago
12

The director of a local ballet company needs to print the programs for the next performance. Janet’s Print Shop charges $0.25 pe

r program plus a $35 set-up fee. The Printing Press charges $0.15 per program plus a $50 set-up fee.
Which printing company offers the better deal if 200 programs are printed?
Mathematics
2 answers:
CaHeK987 [17]3 years ago
7 0

Answer:

The Printing Press

Step-by-step explanation:

200 x 0.15 = 30

30 + 50 = £80

200 x 0.25 = 50

50 + 35 = 85

Gnesinka [82]3 years ago
6 0

Answer:

or this case we find the expression that represents the cost of each printing press.

x: Be the variable that represents the number of programs to print

So:

Cost 1: Janet’s Print Shop

Cost 2: The Printing Press

If we want to find the number of programs for which the costs are the same, then we equate both equations:

Thus, for 150 programs the cost is the same.

Answer:

For 150 programs the cost is the same.

Step-by-step explanation:

Hope this helps you!!!!!!! :D

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If you add one-third of a number to the number itself, you get 16.what is the number?
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One third of 12 is 4.

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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
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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

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Find an angle between 0 and 360     that is coterminal with 900
Vladimir79 [104]

I'm guessing its 1800.

4 0
3 years ago
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