In an experiment of 50 cars passing was recorded 17 silver cars were recorded what Is the relative frequency of a silver car pas
1 answer:
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To solve this we are going to use the future value of annuity ordinary formula: ![FV=P[ \frac{(1+ \frac{r}{n} )^{kt} -1}{ \frac{r}{n} } ]](https://tex.z-dn.net/?f=FV%3DP%5B%20%5Cfrac%7B%281%2B%20%5Cfrac%7Br%7D%7Bn%7D%20%29%5E%7Bkt%7D%20-1%7D%7B%20%5Cfrac%7Br%7D%7Bn%7D%20%7D%20%5D)
where
is the future value
is the periodic payment
is the interest rate in decimal form
is the number of times the interest is compounded per year
is the number of payments per year
is the number of years
We know for our problem that
and 
. To convert the interest rate to decimal form, we are going to divide the rate by 100%:
Since the deposit is made semiannually, it is made 2 times per year, so
.
Since the type of the annuity is ordinary, payments are made at the end of each period, and we know that we have 2 periods, so
.
Lets replace the values in our formula:
![FV=6200[ \frac{(1+ \frac{0.06}{2} )^{(2)(5)} -1}{ \frac{0.06}{2} } ]](https://tex.z-dn.net/?f=FV%3D6200%5B%20%5Cfrac%7B%281%2B%20%5Cfrac%7B0.06%7D%7B2%7D%20%29%5E%7B%282%29%285%29%7D%20-1%7D%7B%20%5Cfrac%7B0.06%7D%7B2%7D%20%7D%20%5D)
We can conclude that the correct answer is <span>$71,076.06</span>
where
We know for our problem that
Since the deposit is made semiannually, it is made 2 times per year, so
Since the type of the annuity is ordinary, payments are made at the end of each period, and we know that we have 2 periods, so
Lets replace the values in our formula:
We can conclude that the correct answer is <span>$71,076.06</span>