In the diagram, the distance from A to B is 4 inches. Find the actual distance if the scale is 1 in:6 mi
1 answer:
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Hey there,
So explain and to illustrate (how
is the midpoint of 
and the length of is 23.8 what is the length of 
<span> A-----------------B----------------C
</span>
So,
in the problem here is to be considered the (Whole area of this line). is the whole thing.
So, what we do in this problem, we do . . .
23.8 ÷ 2 because we need to find what is, so we will find the one half of it, which in this illustration is to be considered
23.8 ÷ 2 =
Hope this helps.
~Jurgen
So explain and to illustrate (how
<span> A-----------------B----------------C
</span>
So,
So, what we do in this problem, we do . . .
23.8 ÷ 2 because we need to find what is
23.8 ÷ 2 =
Hope this helps.
~Jurgen
Answer:
a)0.62% probability that total sick leave for next month will be less than 150 hours.
b) 225.6 hours should be budgeted for sick leave if that amount is to be exceeded with a probability of only 0.10.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
a.Find the probability that total sick leave for next month will be less than 150 hours.
This probability is the pvalue of Z when X = 150. So:
has a pvalue of 0.0062.
So there is a 0.62% probability that total sick leave for next month will be less than 150 hours.
b.In planning schedules for next month, how much time should be budgeted for sick leave if that amount is to be exceeded with a probability of only 0.10.
This is the value of X when Z has a pvalue of 0.90. So
225.6 hours should be budgeted for sick leave if that amount is to be exceeded with a probability of only 0.10.