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
2 answers:
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.
Absolute difference written diameter of 58.2 mm and average diameter is
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.
Therefore 2 standard deviations a ball does bearing with a diameter of 58.2 mm differ from the mean.
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Answer:
the angle between their paths is <em>100.8°</em>
Step-by-step explanation:
From the given information, you can construct a triangle, just like the one in the figure.
We will use the <em>Cosine Rule</em> which is:
c² = b² + a² - 2 b c cos(θ)
where
- c = 16 miles
- b = 8 miles
- a = 12 miles
Therefore,
2 b c cos(θ) = b² + a² - c²
cos(θ) = (b² + a² - c²) / 2 b c
θ = cos⁻¹( (b² + a² - c²) / (2 b c) )
θ = cos⁻¹( (8² + 12² - 16²) / 2(8)(16) )
<em>θ = 100.8°</em>
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Therefore, the angle between their paths is <em>100.8°</em>
