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Ksju [112]
2 years ago
14

A shipment of racquetballs with a mean diameter of 60 mm and a standard deviation of 0.9 mm is normally distributed. By how many

standard deviations does a ball bearing with a diameter of 58.2 mm differ from the mean?
Mathematics
2 answers:
asambeis [7]2 years ago
8 0
Mean 60 - 2(0.9) = 58.2
58.2 is 2 standard deviations from the mean.
tigry1 [53]2 years ago
8 0

Answer:

By 2 standard deviations a ball does bearing with a diameter of 58.2 mm differ from the mean.

Step-by-step explanation:

It is given that a shipment of racquetballs with a mean diameter of 60 mm and a standard deviation of 0.9 mm is normally distributed.

Mean=60

\text{Standard deviation}=0.9

Absolute difference written diameter of 58.2 mm and average diameter is

|58.2-60|=1.8

Divide the difference by standard deviation (i.e.,0.9), to find the by how many standard deviations does a ball bearing with a diameter of 58.2 mm differ from the mean.

\frac{1.8}{0.9}=2

Therefore 2 standard deviations a ball does bearing with a diameter of 58.2 mm differ from the mean.

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At time t=0 hours, a tank contains 3000 litres of water. Water leaks from the tank. At the end of every hour there is x% less wa
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Answer:

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Step-by-step explanation:

At time, t = 0, Volume of Water = 3,000 litres

After t hours , the Volume of water in the tank = Vt

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After 1 hour, volume of water left Vt = V1

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After 2 hours, Volume of water left is V2

V2 = (3,000 - 30 x) - (3,000 - 30 x) × (x/100)

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(3000 - 30x)(100 - x) = 288120

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Factorizing:

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Step 2: Solving for k

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V1 = 3000 - 30 × 2

V1 = 3000 - 60

V1 = 2940

K = 2940/3000

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