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OverLord2011 [107]
4 years ago
9

The standard diameter of a golf ball is 42.67 mm. A golf ball factory does quality control on the balls it manufactures. Golf ba

lls are randomly measured to ensure the correct size. One day, an inspector decides to stop production if the discrepancy in diameter is more than 0.002 mm. What is the range of acceptable values?
Mathematics
2 answers:
Free_Kalibri [48]4 years ago
8 0

Answer:

C) f(x)=|42.67-x|.

Step-by-step explanation:

Production will be stopped if the difference in diameter is greater than 0.002 mm.

This difference can mean that the golf ball has a diameter more than 0.002 larger than 42.67 mm, or a diameter more than 0.002 smaller than 42.67. This is the reason for using an absolute value function.

 Absolute value represents the distance a number is from 0. For this function, we would want only the values where the function is less than 0.002.

SSSSS [86.1K]4 years ago
5 0

Answer:

42.668 mm to 42.672 mm.

Step-by-step explanation:

According to given situation, Production will be stopped if the discrepancy in diameter is greater than 0.002 mm.

The difference in the limits are,

42.67 - 0.002 = 42.668

42.67 + 0.002 = 42.672

Therefore, the range of acceptable values are 42.668 mm to 42.672 mm

0.002 larger than 42.67 mm, or a diameter more than 0.002 smaller than 42.67. This is the reason for using an absolute value function.

 Absolute value represents the distance a number is from 0. For this function, we would want only the values where the function is less than 0.002.

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Historical data indicates that only 20% of cable customers are willing to switch companies. If a binomial process is assumed, th
Natasha_Volkova [10]

Answer:

a. The probability is 0.735

b. The probability is 0.6296

c. The probability is 0

Step-by-step explanation:

If we assume a binomial process, the probability that x customer are willing to switch companies is:

P(x)=nCx*p^{x}*(1-p)^{n-x}

nCx is calculated as:

nCx=\frac{n!}{x!(n-x)!}

Where n are the 20 cable customers and p is the probability 0.2 that the cable customer are willing to switch companies.

Then, P(x) is:

P(x)=20Cx*0.2^{x}*(1-0.2)^{20-x}

The probability that between 2 and 5 (inclusive) customers are willing to switch companies is:

P(2≤x≤5) = P(2) + P(3) + P(4) + P(5)

Where P(2), P(3), P(4) and P(5) are equal to:

P(2)=20C2*0.2^{2}*(1-0.2)^{20-2}=0.1369

P(3)=20C3*0.2^{3}*(1-0.2)^{20-3}=0.2054

P(4)=20C4*0.2^{4}*(1-0.2)^{20-4}=0.2182

P(5)=20C5*0.2^{5}*(1-0.2)^{20-5}=0.1745

So, P(2≤x≤5) is:

P(2≤x≤5) = 0.1369 + 0.2054 + 0.2182 + 0.1745 = 0.735

At the same way, the probability that less than 5 customers are willing to switch is:

P(x<5)=P(0)+P(1)+P(2)+P(3)+P(4)

P(x<5)=0.6296

Finally, the probability that more than 16 customers are willing to switch is:

P(x>16)=P(17)+P(18)+P(19)+P(20)

P(x>16)=0

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