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2 years ago

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1.What is the principal which would amount to $6 900 in 3 years at 5% simple interest per annum?Required to answer. Single choic

e.
(1 Point)

a.) $6 885

b.) $6 555

c.) $6 000

d.) $5 865

1 answer:

Jobisdone [24]2 years ago

8 0

Answer:

c

Step-by-step explanation:

I just looked up the question

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The table below shows the ages of employees over 50 at a company and the amount of money they have in their retirement accounts.

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3 years ago

NarStor, a computer disk drive manufacturer, claims that the median time until failure for their hard drives is more than 14,400

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Answer:

The test statistics is $t = -1.727$

Step-by-step explanation:

From the question we are told that

The data given is

330 620 1870 2410 4620 6396 7822 81028309 12882 14419 16092 18384 20916 23812 25814

The population mean is $\mu = 14400$

The sample size is n = 16

The null hypothesis is $\mu \le 14400$

The alternative hypothesis is $H_a : \mu > 14400$

The sample mean is mathematically evaluated as

$\= x = \frac{\sum x_i}{n}$

So

$\= x = \frac{330+ 620+ 1870 +2410+ 4620+ 6396+ 7822+ 8102+8309+ 12882+ 14419+ 16092+ 18384 +20916+ 23812+ 25814 }{16}$

=> $\= x = 10799.9$

The standard deviation is mathematically represented as

$\sigma =\sqrt{\frac{ \sum (x_i - \=x)^2}{n}}$

So

$\sigma =\sqrt{\frac{(330- 10799.9)^2 + (620- 10799.9)^2+ (1870- 10799.9)^2 +(2410- 10799.9)^2 + (4620- 10799.9)^2 +(6396- 10799.9)^2 +(7822- 10799.9)^2 }{16}} \ ..$

$..\sqrt{ \frac{(8102 - 10799.9)^2 +(8309 - 10799.9)^2 + (12882 - 10799.9)^2 + (14419 - 10799.9)^2 + (16092 - 10799.9)^2 + (18384 - 10799.9)^2 +(20916 - 10799.9)^2 }{16}} \ ...$

$\ ... \sqrt{\frac{(23812 - 10799.9)^2 +(25814 - 10799.9)^2 }{16}}$

=> $\sigma = 8340$

Generally the test statistic is mathematically represented as

$t = \frac{10799.9- 14400}{ \frac{8340}{\sqrt{16} } }$

$t = -1.727$

From the z-table the p-value is

$p-value = P(Z > t) = P(Z > -1.727) = 0.95792$

From the values obtained we see that

$p-value > \alpha$ so we fail to reject the null hypothesis

Which implies that the claim of the NarStor is wrong

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