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kherson [118]
3 years ago
9

A printing company is considering buying a new printing press. It has quotes from two different companies. Company A offers a pr

ess for $9,500 plus $75 per month for a repair contract that covers unlimited repairs. Company B offers a press for $10,500 and charges $150 per repair.
The table gives the probabilities of the numbers of repairs to the printing press offered by company
b. Number of Repairs
0 1 2 3
Probability
0.32 0.29 0.25 0.14

which offer is more cost-effective?
Mathematics
1 answer:
Ostrovityanka [42]3 years ago
8 0
The complete problem wants to know the more cost-effective choice is the printing company replaces its press machine every four years. With this in mind, if they choose Company A, the total amount that needs to be paid to cover 12(4) = 48 months will be Company A = 9 500 + 48(750) = $13 100 As for Company B, the highest probability of not having any repairs over the four years is 0.32 while there's 0.14 chance that number of repairs will be 3 over the time the contract is applied. This means that the possible amounts to be paid to Company B are Company B = 10 500 + 0(150) = $10 500 Company B = 10 500 + 1(150) = $10 650 Company B = 10 500 + 2(150) = $10 800 Company B = 10 500 + 3(150) = $10 950 This shows that choosing the machine from Company B would be better. Hence, the answer is Company B<span>.</span>
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The distribution of SAT II Math scores is approximately normal with mean 660 and standard deviation 90. The probability that 100
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Using the <em>normal distribution and the central limit theorem</em>, it is found that there is a 0.1335 = 13.35% probability that 100 randomly selected students will have a mean SAT II Math score greater than 670.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

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  • After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
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In this problem:

  • The mean is of 660, hence \mu = 660.
  • The standard deviation is of 90, hence \sigma = 90.
  • A sample of 100 is taken, hence n = 100, s = \frac{90}{\sqrt{100}} = 9.

The probability that 100 randomly selected students will have a mean SAT II Math score greater than 670 is <u>1 subtracted by the p-value of Z when X = 670</u>, hence:

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{670 - 660}{9}

Z = 1.11

Z = 1.11 has a p-value of 0.8665.

1 - 0.8665 = 0.1335.

0.1335 = 13.35% probability that 100 randomly selected students will have a mean SAT II Math score greater than 670.

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X-4y-8=6/8+-8y-2y<br>5y-3(6-7/8)=5x-4y+3
Ivan
First simplify

x-4y-8=3/4-10y
add 10y both sides and add 8
x+6y=8+3/4
times 4 both sides
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4x+24y=35



5y-3(6-7/8)=5x-4y+3
add 4y both sides
9y-3(6-7/8)=5x+3
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add 18 to both sides and minus 21/8 both sides
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so we got

24y+4x=35
72y-40x=147

multiply first equation by -3 and add to 2nd

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72y<span>-40x=147 +</span>
0y-52x=42


-52x=42
divide both sides by -52
x=-21/26
sub back

24y+4x=35
24y+4(-21/26)=35
using math
y=497/312


the solution is
( \frac{-21}{26}, \frac{497}{312}  )
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