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shepuryov [24]
3 years ago
7

Jeanne Crawford had $9,675.95 deposited in an account paying

Mathematics
1 answer:
Anni [7]3 years ago
8 0

Answer:

a. $10,943.30

b. $1,267.35

Step-by-step explanation:

P = $9,675.95

r = 6.25% = 0.0625

t = Compounded Semiannually = 2

a. Amount after 2 Years

n = 2

A = P [1 + (r / n)]^nt

A = $9,675.95 [1 + (0.0625 / 2)]² ˣ ²

A = $9,675.95 [1 + 0.03125]⁴

A = $9,675.95 [1.03125]⁴

A = $9,675.95 x 1.130982

A = $10,943.30

b. Compound Interest

Compound Interest = Final Amount - Principal Amount

Compound Interest = $10,943.30 - $9,675.95

Compound Interest = $1,267.35

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1. 24% of 800<br> 75% of 1000
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Step-by-step explanation:

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3 years ago
A recent survey indicated that the average amount spent for breakfast by business managers was $9.33 with a standard deviation o
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Answer:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

Replacing the info given we got:

t=\frac{9.41-9.33}{\frac{0.24}{\sqrt{81}}}=3    

Step-by-step explanation:

Information given

\bar X=9.41 represent the sample mean

s=0.24 represent the sample standard deviation

n=81 sample size  

\mu_o =9.33 represent the value to verify

\alpha=0.05 represent the significance level

t would represent the statistic

p_v represent the p value

Hypothesis to test

We want to test if the true mean is higher than 9.33, the system of hypothesis would be:  

Null hypothesis:\mu \leq 9.33  

Alternative hypothesis:\mu > 9.33  

The statistic is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

Replacing the info given we got:

t=\frac{9.41-9.33}{\frac{0.24}{\sqrt{81}}}=3    

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