keeping in mind that 4 months is not even a year, since there are 12 months in a year, 4 months is then 4/12 years.
let's assume is simple interest.
![\bf ~~~~~~ \textit{Simple Interest Earned Amount} \\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill & \$34300\\ r=rate\to 3.5\%\to \frac{3.5}{100}\dotfill &0.035\\ t=years\to \frac{4}{12}\dotfill &\frac{1}{3} \end{cases} \\\\\\ A=34300\left[ 1+(0.035)\left( \frac{1}{3} \right) \right]\implies A= 34300(1.011\overline{6})\implies A=34700.1\overline{6}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%20%5Ctextit%7BSimple%20Interest%20Earned%20Amount%7D%20%5C%5C%5C%5C%20A%3DP%281%2Brt%29%5Cqquad%20%5Cbegin%7Bcases%7D%20A%3D%5Ctextit%7Baccumulated%20amount%7D%5C%5C%20P%3D%5Ctextit%7Boriginal%20amount%20deposited%7D%5Cdotfill%20%26%20%5C%2434300%5C%5C%20r%3Drate%5Cto%203.5%5C%25%5Cto%20%5Cfrac%7B3.5%7D%7B100%7D%5Cdotfill%20%260.035%5C%5C%20t%3Dyears%5Cto%20%5Cfrac%7B4%7D%7B12%7D%5Cdotfill%20%26%5Cfrac%7B1%7D%7B3%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20A%3D34300%5Cleft%5B%201%2B%280.035%29%5Cleft%28%20%5Cfrac%7B1%7D%7B3%7D%20%5Cright%29%20%5Cright%5D%5Cimplies%20A%3D%2034300%281.011%5Coverline%7B6%7D%29%5Cimplies%20A%3D34700.1%5Coverline%7B6%7D)
Answer:
The z-score for the trainee is of 2.
Step-by-step explanation:
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
The mean of the test scores is 72 with a standard deviation of 5.
This means that 
Find the z-score for a trainee, given a score of 82.
This is Z when X = 82. So



The z-score for the trainee is of 2.
Answer:
so the answer would points on al ine you got to make sure you look at the 5
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