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3 years ago

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2. A bank representative studies compound interest, so she can better serve

1 answer:

Kaylis [27]3 years ago

4 0

Answer:

Step-by-step explanation:

Given that a bank representative studies compound interest, so she can better serve customers. She analyzes what happens when $2,000 earns interest several different ways at a rate of 2% for 3 years.

a) the interest if it is computed using simple interest. 12.00

= $\frac{2000(3)(12)}{100} \\=720$ dollars

b) the interest if it is compounded annually.

= $2000(1.12)^3-2000\\= 809.86$ dollars

c) the interest if it is compounded semiannually

= $2000(1.06)^6 -2000\\= 837.04$

d) the interest if it is compounded quarterly.

= $2000(1.03)^12 -2000\\= 851.52$

e) the interest if it is compounded monthly.

= $2000(1.01)^24 -2000\\= 861.54$

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Twenty-one boys and eight girls go on a camping trip. In how many ways can a group of six be selected to gather firewood, given

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(a) there are 8C2 = 28 ways of picking 2 girls from 8
And there are 21C4 = 5985 ways of picking 4 boys

Required number of ways for 2g / 4b = 28 * 5985 = 167,580

(b) at least 2 girls means combinations of 2g/4b , 3g,3b , 4g/2b , 5g 1b or
6 girls.

2g/4b = 167,580 ways
3g/3b = 8C3 * 21C3 = 56 * 1330 = 74,480
4g/2b = 8C4* 21C2 = 70 * 210 = 14,700
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3 years ago

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Answer:

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2 years ago

[1 -4 3 5]+[-2 6 -2 4]

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Step-by-step explanation:

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2 years ago

Consider the following hypothesis test. H0: μ ≥ 55 Ha: μ < 55 A sample of 36 is used. Identify the p-value and state your con

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Answer:

Step-by-step explanation:

Given that:

$H_o: \mu \ge 55 \ \\ \\ H_1 : \mu < 55$

(a) For x = 54 and s = 5.3

The test statistics can be computed as:

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

$Z = \dfrac{54-55}{\dfrac{5.3}{\sqrt{36}} }$

$Z = \dfrac{-1}{\dfrac{5.3}{6} }$

Z = -1.132

degree of freedom = n - 1

= 36 - 1

= 35

Using the Excel Formula:

P-Value = T.DIST(-1.132,35,1) = 0.1326

Decision: p-value is greater than significance level; do not reject $\mathbf{H_o}$

$Conclusion: \ There \ is \ insufficient \ evidence \ to \ conclude \ that \$ $\mu < 55$

b

For x = 53 and s = 4.6

The test statistics can be computed as:

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

$Z = \dfrac{53-55}{\dfrac{4.6}{\sqrt{36}} }$

$Z = \dfrac{-2}{\dfrac{4.6}{6} }$

Z = -2.6087

degree of freedom = n - 1

= 36 - 1

= 35

Using the Excel Formula:

P-Value = T.DIST(-2.6087,35,1) =0.0066

Decision: p-value is < significance level; we reject the null hypothesis.

$Conclusion: \ There \ is \ sufficient \ evidence \ to \ conclude \ that$ $\mu < 55$

c)

For x = 56 and s = 5.0

The test statistics can be computed as:

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

$Z = \dfrac{56-55}{\dfrac{5.0}{\sqrt{36}} }$

$Z = \dfrac{56-55}{\dfrac{5.0}{\sqrt{36}} }$

Z = 1.2

degree of freedom = n - 1

= 36 - 1

= 35

Using the Excel Formula:

P-Value = T.DIST(1.2,35,1) = 0.88009

Decision: p-value is greater than significance level; do not reject $\mathbf{H_o}$

$Conclusion: \ There \ is \ insufficient \ evidence \ to \ conclude \ that \$ $\mu < 55$

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Answer:

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