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Answer:
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Answer:
After 10 years, she will has $96 in her bank.
Step-by-step explanation:
It is given that Andrea's saving account is $80 and earns 2% interest per year as a <em>S</em><em>i</em><em>m</em><em>p</em><em>l</em><em>e</em><em> </em><em>I</em><em>n</em><em>t</em><em>e</em><em>r</em><em>e</em><em>s</em><em>t</em><em> </em>(Not Compounded). Using simple interest formula, Interest = (P×R×T)/100 where <em>P</em> is the <em>principal</em>, <em>R</em> is the <em>interest rate</em> and <em>T</em> is <em>number of years</em><em> </em>:
P = $80
R = 2%
T = 10 years
It is given that the interest amount is $16. So the total amount she has after 10 years in the bank is $96 :
interest amount = $16
principal = $80
total = $16 + $80
= $96
Answer:
They should guarantee the lifetime of their batteries for 32 months.
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.
Mean of 36 months and a standard deviation of 2 months.
This means that
If the company wants to replace no more than 2% of all batteries, for how many months should they guarantee the lifetime of their batteries?
The guarantee should be the 2th percentile of lengths, which is X when Z has a pvalue of 0.02. So X when Z = -2.054.
Rounding to the closest month, 32.
They should guarantee the lifetime of their batteries for 32 months.