The temperature went from 83 at noon to 72 at midnight. What is the percent of decrease in the temperature from noon to midnight
2 answers:
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This would help you alot.
That's the diagram..
(12×3) + 3(0.67×7) = 50.07inches²
≈ 50inches²
hope this helps
make it the brainliest if it's correct....
(12×3) + 3(0.67×7) = 50.07inches²
≈ 50inches²
hope this helps
make it the brainliest if it's correct....
Answer:
ans=13.59%
Step-by-step explanation:
The 68-95-99.7 rule states that, when X is an observation from a random bell-shaped (normally distributed) value with mean and standard deviation
, we have these following probabilities
In our problem, we have that:
The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 53 months and a standard deviation of 11 months
So
So:
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What is the approximate percentage of cars that remain in service between 64 and 75 months?
Between 64 and 75 minutes is between one and two standard deviations above the mean.
We have subtracted by
is the percentage of cars that remain in service between one and two standard deviation, both above and below the mean.
To find just the percentage above the mean, we divide this value by 2
So:
The approximate percentage of cars that remain in service between 64 and 75 months is 13.59%.