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VMariaS [17]
3 years ago
13

A company finds it can produce 5 heaters for $1750, while producing 10 heaters cost $3000. Express the cost, y, as a linear func

tion of the number of heaters, x.
Mathematics
1 answer:
fiasKO [112]3 years ago
8 0

The cost function is linear, and as such has a slope and a y-intercept.

Two points on this line are (5, $1750) and (10, $3000).  As the number of heaters produced increases from 5 to 10, the production costs increase by $1250.  Thus, the slope of this line is m = $1250/5, or m = $250/unit.

Inserting known data into the slope-intercept form of the equation of a straight line, we get:

$3000 = ($250/unit)(10) + b.  Then $500 = b, and the cost function is:

y = ($250/unit)(x) + $500.

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Step-by-step explanation:

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Suppose that a basketball player can score on a particular shot with probability .3. Use the central limit theorem to find the a
Rom4ik [11]

Answer:

(a) The probability that the number of successes is at most 5 is 0.1379.

(b) The probability that the number of successes is at most 5 is 0.1379.

(c) The probability that the number of successes is at most 5 is 0.1379.

(d) The probability that the number of successes is at most 11 is 0.9357.

→ All the exact probabilities are more than the approximated probability.

Step-by-step explanation:

Let <em>S</em> = a basketball player scores a shot.

The probability that a basketball player scores a shot is, P (S) = <em>p</em> = 0.30.

The number of sample selected is, <em>n</em> = 25.

The random variable S\sim Bin(25,0.30)

According to the central limit theorem if the sample taken from an unknown population is large then the sampling distribution of the sample proportion (\hat p) follows a normal distribution.

The mean of the the sampling distribution of the sample proportion is: E(\hat p)=p=0.30

The standard deviation of the the sampling distribution of the sample proportion is:

SD(\hat p)=\sqrt{\frac{ p(1- p)}{n} }=\sqrt{\frac{ 0.30(1-0.30)}{25} }=0.092

(a)

Compute the probability that the number of successes is at most 5 as follows:

The probability of 5 successes is: p=\frac{5}{25} =0.20

P(\hat p\leq 0.20)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq  \frac{0.20-0.30}{0.092} )\\=P(Z\leq -1.087)\\=1-P(Z

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 5 is 0.1379.

The exact probability that the number of successes is at most 5 is:

P(S\leq 5)={25\choose 5}(0.30)^{5}91-0.30)^{25-5}=0.1935

The exact probability is more than the approximated probability.

(b)

Compute the probability that the number of successes is at most 7 as follows:

The probability of 5 successes is: p=\frac{7}{25} =0.28

P(\hat p\leq 0.28)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq  \frac{0.28-0.30}{0.092} )\\=P(Z\leq -0.2174)\\=1-P(Z

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 7 is 0.4129.

The exact probability that the number of successes is at most 7 is:

P(S\leq 57)={25\choose 7}(0.30)^{7}91-0.30)^{25-7}=0.5118

The exact probability is more than the approximated probability.

(c)

Compute the probability that the number of successes is at most 9 as follows:

The probability of 5 successes is: p=\frac{9}{25} =0.36

P(\hat p\leq 0.36)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq  \frac{0.36-0.30}{0.092} )\\=P(Z\leq 0.6522)\\=0.7422

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 9 is 0.7422.

The exact probability that the number of successes is at most 9 is:

P(S\leq 9)={25\choose 9}(0.30)^{9}91-0.30)^{25-9}=0.8106

The exact probability is more than the approximated probability.

(d)

Compute the probability that the number of successes is at most 11 as follows:

The probability of 5 successes is: p=\frac{11}{25} =0.44

P(\hat p\leq 0.44)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq  \frac{0.44-0.30}{0.092} )\\=P(Z\leq 1.522)\\=0.9357

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 11 is 0.9357.

The exact probability that the number of successes is at most 11 is:

P(S\leq 11)={25\choose 11}(0.30)^{11}91-0.30)^{25-11}=0.9558

The exact probability is more than the approximated probability.

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Is it the green?<br><br>Which line has a slope of 2?
OleMash [197]

Answer:

With all the lines, I may be wrong. But I'm confident that line B has a slope of 2.

This is because it rises up 2 up the y axis for every one unit on the x axis.

8 0
3 years ago
Read 2 more answers
Annie’s change purse contained 5 quarters 8 dimes and 8 nickels if Annie chose 2 coin randomly without replacing them in her pur
Bogdan [553]
<h3>Answer:</h3>

2/15

<h3>Explanation:</h3>

There are 8C2 = 28 ways to choose 2 dimes from the 8 dimes in Annie's purse. There are 21C2 = 210 ways to choose 2 coins from the 21 coins in Annie's purse.

Of the 210 ways to choose 2 coins, 28 of the choices will result in 2 dimes being chosen. The probability of choosing 2 dimes is 28/210 = 2/15.

_____

<em>Comment on nCk</em>

The number of ways to choose k objects from n, when order does not matter, is ...

... n!/(k!(n -k)!)

For the computations above, we have ...

... 8C2 = 8·7/(2·1) = 28

... 21C2 = 21·20/(2·1) = 210

8 0
2 years ago
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