alienxangg91
29.07.2020 •
Mathematics
2. Suppose that the mean salary in a particular profession is $45,000 with a standard deviation of $1,500. What percentage of people in that profession earn less than $48,000
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Ответ:
93%
Step-by-step explanation:
mean=45,000 standard deviation=2000 value of concern=48,000
We can easily see that since the value of concern (48,000) is GREATER than the mean, we can rule out the last two choices.
There is no possible way a number can be greater than the mean, but less than the 50th percentile.
convert 48,000 into a z-score, which is given as:
(x-mean)/standard deviation
or in this case:
(48000-45000)/2000=1.5
using my z-score table or calculator, I can see that a z-score of 1.5 corresponds to about the 93th percentile
Ответ:
4762 N/m^2
Step-by-step explanation:
Find area.
30x5.6=168
Convert to m^2
168cm^2=0.0168m^2
Use pressure formula to get answer
P=F/A
P=80/0.0168
P=4762 N/m^2