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Sedbober [7]
3 years ago
5

A 20,000 deposit earns 4.8 interest for 4 years . What is the final account balance if the interest is compounded semiannually ?

Quarterly? Monthly ?
Mathematics
1 answer:
Bad White [126]3 years ago
4 0

Answer:

Semi-annually: A = $24 178.51

Quarterly:         A = $24 205.73

Monthly:           A = $24 224.13

Step-by-step explanation:

The formula for compound interest is

A = P(1 + r)ⁿ

A. Compounded semi-annually

Data:

P = $20 000

APR  = 4.8 %

t = 4 yr

Calculations:

n = 4 × 2 = 8

r = 0.048/2  = 0.024

A = 20 000(1+ 0.024)⁸

  = 20 000 × 1.024⁸

  = 20 000 × 1.208 926

  = $24 178.51

B. Compounded Quarterly

n = 4 × 4 = 16

r = 0.048/4  = 0.012

A = 20 000(1+ 0.012)¹⁶

  = 20 000 × 1.012¹⁶

  = 20 000 × 1.210 286

  = $24 205.73

C. Compounded monthly

n = 4 × 12 = 48

r = 0.048/12  = 0.004

A = 20 000(1+ 0.004)⁴⁸

  = 20 000 × 1.004⁴⁸

  = 20 000 × 1.211 207

  = $24 224.13

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In tests of a computer component, it is found that the mean time between failures is 937 hours. A modification is made which is
VladimirAG [237]

Answer:

Null hypothesis is \mathbf {H_o: \mu > 937}

Alternative hypothesis is \mathbf {H_a: \mu < 937}

Test Statistics z = 2.65

CONCLUSION:

Since test statistics is greater than  critical value; we reject the null hypothesis. Thus, there is sufficient evidence to support the claim that the modified components have a longer mean time between failures.

P- value = 0.004025

Step-by-step explanation:

Given that:

Mean \overline x = 960 hours

Sample size n = 36

Mean population \mu = 937

Standard deviation \sigma = 52

Given that the mean  time between failures is 937 hours. The objective is to determine if the mean time between failures is greater than 937 hours

Null hypothesis is \mathbf {H_o: \mu > 937}

Alternative hypothesis is \mathbf {H_a: \mu < 937}

Degree of freedom = n-1

Degree of freedom = 36-1

Degree of freedom = 35

The level of significance ∝ = 0.01

SInce the degree of freedom is 35 and the level of significance ∝ = 0.01;

from t-table t(0.99,35), the critical value = 2.438

The test statistics is :

Z = \dfrac{\overline x - \mu }{\dfrac{\sigma}{\sqrt{n}}}

Z = \dfrac{960-937 }{\dfrac{52}{\sqrt{36}}}

Z = \dfrac{23}{8.66}

Z = 2.65

The decision rule is to reject null hypothesis   if  test statistics is greater than  critical value.

CONCLUSION:

Since test statistics is greater than  critical value; we reject the null hypothesis. Thus, there is sufficient evidence to support the claim that the modified components have a longer mean time between failures.

The P-value can be calculated as follows:

find P(z < - 2.65) from normal distribution tables

= 1 - P (z ≤ 2.65)

= 1 - 0.995975     (using the Excel Function: =NORMDIST(z))

= 0.004025

6 0
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Nadia and some friends went to a movie. Their total cost was $30.24 which included fasted of 2.24 write an algebraic expreeeion
alexgriva [62]

Answer:

You can put this solution

30.24 - 2.24 = $28 just tickets

 

$28/x = $/person

or

(30.24 - 2.24)/x

Step-by-step explanation:

6 0
2 years ago
Could someone please help me with this thank you very much
elena-s [515]

Answer:

Answer for #10:

y = 2 \: and \: x = 1

Answer for #1:

x = 2 \: and \: y =  - 1

Step-by-step explanation:

In order to solve a simultaneous equation using the substitution method you have to:

Use one of the equation and substitute it into the other equation, in this case equation <em>B</em><em> </em>was substituted into equation <em>A</em><em>.</em>

<em>equ. \: a = (2x - 3y = 7) \\ equ. \: b = (x = 1 - y)</em>

Therefore since x = 1 - y, equation A would be:

2(1 - y) - 3y = 7 \\ (2 - 2y) - 3y = 7 \\  - 2y - 3y = 7 - 2 \\  - 5y = 5 \\ y =  - 1 \: and \: x = 2

8 0
1 year ago
What is the ◼️ root of 2037?
Ksju [112]
Hello there!

I believe the answer is this.

hope this helps!

5 0
3 years ago
Read 2 more answers
(1/3) to the 4th power
stepan [7]

Answer:

I think the answer might be 0.012

Step-by-step explanation:

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