Help me please , thank you so much
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>
Answer:
The number of individuals included in the summary is 146.
Step-by-step explanation:
The frequency distribution table provided is as follows:
Class Intervals Frequency
100 - 199 24
200 - 299 90
300 - 399 27
400 - 499 1
500 - 599 4
The lower class limit it the smallest value of each class interval.
Lower class limit = {100, 200, 300, 400, 500}
The upper class limit it the highest value of each class interval.
Upper class limit = {199, 299, 399, 499, 599}
The lower class boundaries are the lower class limits decreased by 0.5 and the upper class boundaries are the upper class limits increased by 0.5.
Class boundaries:
99.5 - 199.5
199.5 - 299.5
299.5 - 399.5
399.5 - 499.5
499.5 - 599.5
The class width is the difference between the class boundaries of each class.
Class width = 199.5 - 99.5 = 100
So, the class width is 100.
The midpoints of a class is the average value of the boundaries of a class.
The number of individuals included in the summary is the sum of all frequencies.
Thus, the number of individuals included in the summary is 146.