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myrzilka [38]
3 years ago
6

The lengths of a particular type of metal rod are modeled by the equation r = |–24x + 120|, where r represents the lengths of th

e rod in inches, and x represents the speed at which the rods are produced, in rods per minute. If the manufacturing process is too fast or too slow, the rods will not be long enough. At what speeds will the lengths of the rods be greater than 48 inches?
A. The length of the rods will be greater than 48 inches when x > 3 or x < 7 rods per minute.


B. The length of the rods will be greater than 48 inches when x < 3 or x > 7 rods per minute.


C. The length of the rods will be greater than 48 inches when x > –3 or x < –7 rods per minute.


D. The length of the rods will be greater than 48 inches when 3 < x < 7 rods per minute
Mathematics
1 answer:
Lisa [10]3 years ago
8 0
| -24x + 120| > 48

-24x + 120 > 48
-24x > 48 - 120
-24x > - 72
x < -72/-24
x < 3

-24x + 120 < -48
-24x < -48 - 120
-24x < - 168
x > -168/-24
x > 7

solution is : x < 3 or x > 7...answer B
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Step-by-step explanation:

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Scilla [17]

Answer:

a) P=0.3174

b) P=0.4232

c) P=0.2594

d) The shape of the hypergeometric, in this case, is like a binomial with mean np=1.

Step-by-step explanation:

The appropiate distribution to model this is the hypergeometric distribution:

P(X=x)=\frac{\binom{s}{x}\binom{N-s}{M-x}}{\binom{N}{M}}=\frac{\binom{6}{x}\binom{54}{10-x}}{\binom{60}{10}}

a) What is the probability that none of the questions are essay?

P(X=0)=\frac{\binom{6}{0}\binom{54}{10-0}}{\binom{60}{10}}\\\\P(X=0)=\frac{1*(54!/(10!*44!)}{60!/(10!*50!)} =\frac{2.3931*10^{10}}{7.5394*10^{10}} = 0.3174

b)  What is the probability that at least one is essay?

P(X=1)=\frac{\binom{6}{1}\binom{54}{9}}{\binom{60}{10}}\\\\P(X=1)=\frac{6*(54!/(9!*43!)}{60!/(10!*50!)} =\frac{3.1908*10^{10}}{7.5394*10^{10}} =0.4232

c) What is the probability that two or more are essay?

P(X\geq2)=1-(P(0)+P(1))=1-(0.3174+0.4232)=1-0.7406=0.2594

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