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oee [108]
3 years ago
5

Input

Mathematics
1 answer:
Yuliya22 [10]3 years ago
7 0

Answer: 46

Step-by-step explanation:

Yes !!

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Calculate the a) future value of the annuity due, and b) total interest earned. (From Example 2)
vredina [299]

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.

__

<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.

__

<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.

3 0
2 years ago
I need help with math
Marat540 [252]
The answer is C that’s all I can give to you
6 0
3 years ago
(5x3 + 48x2 + 25x − 18) ÷ (x + 9)
Lorico [155]
(x+1)•(5x-2) hope this helps
5 0
3 years ago
Use the standard norml distribution or the t-distribution to construct a 99% confidence interval for the population mean. Justif
dalvyx [7]

Answer:

E) we will use t- distribution because is un-known,n<30

the confidence interval is (0.0338,0.0392)

Step-by-step explanation:

<u>Step:-1</u>

Given sample size is n = 23<30 mortgage institutions

The mean interest rate 'x' = 0.0365

The standard deviation 'S' = 0.0046

the degree of freedom = n-1 = 23-1=22

99% of confidence intervals t_{0.01} =2.82  (from tabulated value).

The mean value = 0.0365

x±t_{0.01} \frac{S}{\sqrt{n-1} }

0.0365±2.82 \frac{0.0046}{\sqrt{23-1} }

0.0365±2.82 \frac{0.0046}{\sqrt{22} }

0.0365±2.82 \frac{0.0046}{4.690 }

using calculator

0.0365±0.00276

Confidence interval is

(0.0365-0.00276,0.0365+0.00276)

(0.0338,0.0392)

the mean value is lies between in this confidence interval

(0.0338,0.0392).

<u>Answer:-</u>

<u>using t- distribution because is unknown,n<30,and the interest rates are not normally distributed.</u>

4 0
3 years ago
Write the improper fraction as an improper 6 4/13
Contact [7]

Answer:

82/13

Step-by-step explanation:

3 0
3 years ago
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