Answer:
7.64% probability that they spend less than $160 on back-to-college electronics
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:
Probability that they spend less than $160 on back-to-college electronics
This is the pvalue of Z when X = 160. So
has a pvalue of 0.0763
7.64% probability that they spend less than $160 on back-to-college electronics
We have that
the expression 2f + 4f + 2 – 3-------> we can group it (2f+4f)+(2-3)
(2f+4f)+(2-3)=(6f-1)
therefore
the expression [2f + 4f + 2 – 3] is equivalent to [6f-1]
if the expression [6f-1] for f=3 is 17
then
the expression [2f + 4f + 2 – 3] for f=3 is also 17
<span>let's check it
</span>[2*3 + 4*3 + 2 – 3]--------> [6+12+2-3]=[20-3]=17------> is ok
Answer:
116.33
Step-by-step explanation:
3.59
Or 3.593
Or 3.6
It matters what spot u want it reduced to