Answer:
Step-by-step explanation:
we know that
The compound interest formula is equal to
where
A is the Final Investment Value
P is the Principal amount of money to be invested
r is the rate of interest in decimal
t is Number of Time Periods
n is the number of times interest is compounded per year
in this problem we have
substitute in the formula above
Answer:50%
Step-by-step explanation: It doesn't matter how many questions there are because for each question, you have a 50% chance of getting it right. So, the probability is 50%.
Could you please formulate you answer better? So we can get a bette understanding
Answer:
ans=13.59%
Step-by-step explanation:
The 68-95-99.7 rule states that, when X is an observation from a random bell-shaped (normally distributed) value with mean
and standard deviation
, we have these following probabilities



In our problem, we have that:
The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 53 months and a standard deviation of 11 months
So 
So:



-----------



-----
What is the approximate percentage of cars that remain in service between 64 and 75 months?
Between 64 and 75 minutes is between one and two standard deviations above the mean.
We have
subtracted by
is the percentage of cars that remain in service between one and two standard deviation, both above and below the mean.
To find just the percentage above the mean, we divide this value by 2
So:

The approximate percentage of cars that remain in service between 64 and 75 months is 13.59%.
Answer:
Step-by-step explanation:
4