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2 years ago

Compute the geometric mean for the following data on growth factors of an investment for 10 years.

Mathematics

1 answer:

Brrunno [24]2 years ago

4 0

The geometric mean for the above set of data given is: 1.1475 (Option B). See the explanation below.

<h3>What is Geometric Mean?</h3>

The geometric mean can be defined as the average rate of return of a set of values obtained by multiplying the terms together.

Geometric mean is best suited for series with serial correlation, which is notably true for investment portfolios.

<h3>What is the computation justifying the above answer?
</h3>

The formula for geometric mean is given as:

$\left(\prod _{i=1}^{n}x_{i}\right)^{\frac {1}{n}}={\sqrt[{n}]{x_{1}x_{2}\cdots x_{n}}}$

Hence,

The geometric Mean =

1.0147506490423

$\approx$ 1.1475

Learn more about Geometric Mean:
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3 years ago

Given the domain {-2, 1, 5}, what is the range for the relation 4x + y = 3?

steposvetlana [31]

Answer:

Step-by-step explanation:

Domain is the set of all possible inputs (x) for the function f(x)

The range is the set of all possible values for that domain

This function is 4x + y = 3 which can be re-written as y = 3-4x

Plug in each of the domain values and you will get the range

For -2 : y = 3-4(-2) = 3 + 8 = 11

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2 years ago

An experiment was performed to compare the abrasive wear of two different laminated materials. Twelve pieces of material 1 were

vladimir1956 [14]

Answer:

The calculated value Z= 4.8389> 1.96 at 0.05 level of significance.

The null hypothesis is rejected.

There is significance difference between that the abrasive wear of material 1 not exceeds that of material 2 by more than 2 units

Step-by-step explanation:

Given data the samples of material 1 gave an average (coded) wear of 85 units with a sample standard deviation of 4

Mean of the first sample x₁⁻ =85

standard deviation of the first sample S₁ = 4

Given data the samples of material 2 gave an average of 81 and a sample standard deviation of 5.

Mean of the first sample x₂⁻ =81

standard deviation of the first sample S₂ = 5

<u>Null hypothesis: H₀:</u> there is no significance difference between that the abrasive wear of material 1 exceeds that of material 2 by more than 2 units

<u>Alternative hypothesis :H₁: </u>there is significance difference between that the abrasive wear of material 1 exceeds that of material 2 by more than 2 units

Assume the populations to be approximately normal with equal variances.σ₁² =σ₂²

The test statistic

$Z= \frac{x_{1} -x_{2} }{\sqrt{\frac{S^2_{1} }{n_{1} } +\frac{S^2_{2} }{n_{2} } } }$

Given n₁=n₂=60.

$Z= \frac{85-81 }{\sqrt{\frac{(4)^2 }{60 } +\frac{5^2 }{60} } }$

On calculation, we get

Z = $\frac{4}{\sqrt{0.6833} }$

z = 4.8389

The tabulated value Z =1.96 at 0.05 level of significance.

The calculated value Z= 4.8389> 1.96 at 0.05 level of significance.

The null hypothesis is rejected.

Conclusion:-

there is significance difference between that the abrasive wear of material 1 not exceeds that of material 2 by more than 2 units.

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3 years ago

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2 years ago