Answer:
- value: $66,184.15
- interest: $6,184.15
Step-by-step explanation:
The future value can be computed using the formula for an annuity due. It can also be found using any of a variety of calculators, apps, or spreadsheets.
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<h3>formula</h3>
The formula for the value of an annuity due with payment P, interest rate r, compounded n times per year for t years is ...
FV = P(1 +r/n)((1 +r/n)^(nt) -1)/(r/n)
FV = 5000(1 +0.06/4)((1 +0.06/4)^(4·3) -1)/(0.06/4) ≈ 66,184.148
FV ≈ 66,184.15
<h3>calculator</h3>
The attached calculator screenshot shows the same result. The calculator needs to have the begin/end flag set to "begin" for the annuity due calculation.
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<h3>a) </h3>
The future value of the annuity due is $66,184.15.
<h3>b)</h3>
The total interest earned is the difference between the total of deposits and the future value:
$66,184.15 -(12)(5000) = 6,184.15
A total of $6,184.15 in interest was earned by the annuity.
Answer:
exponential decay
Step-by-step explanation:
1=51
2=34
3= 95 (i think)
4=38
Answer:
a
The null hypothesis is 
The alternative hypothesis is 
b

c

d
The decision rule is
Reject the null hypothesis
e
There is sufficient evidence to support the researchers claim
Step-by-step explanation:
From the question we are told that
The first sample size is 
The sample variance for elementary school is 
The second sample size is 
The sample variance for the secondary school is 
The significance level is 
The null hypothesis is 
The alternative hypothesis is 
Generally from the F statistics table the critical value of
at first and second degree of freedom
and
is

Generally the test statistics is mathematically represented as

=> 
=> 
Generally from the value obtained we see that
Hence
The decision rule is
Reject the null hypothesis
The conclusion is
There is sufficient evidence to support the researchers claim