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1 year ago

Calculate the Rate of Return on an initial investment of $6,000 valued at $6,500.

Mathematics

2 answers:

bezimeni [28]1 year ago

8 0

Answer:

8.3%

Step-by-step explanation:

Tju [1.3M]1 year ago

7 0

Answer:

8.3%

Step-by-step explanation:

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Read 2 more answers

What is the length of line segment EF if DE is 6ft and DF is 11ft and angle FDE is 40 degrees

Anna [14]

Answer:

The length of EF = 7.48 feet

Step-by-step explanation:

* Lets consider these tree segment formed triangle DEF

- We have the length of two sides and the measure of the

including angle between these two sides

* So we can use the cos Rule to find the length of the third side

- The cos rule ⇒ a² = b² + c² - 2bc cosA

# a is the side opposite to angle A

# b is the side opposite to angle B

# c is the side opposite to angle C

# Angle A is the including angle between b and c

* In the problem

∵ DE = 6 feet

∵ DF = 11 feet

∵ m∠FDE = 40° ⇒ including angle between DE and DF

and opposite to EF

- By using cos Rule

∴ (EF)² = (DE)² + (DF)² - 2(DE)(DF) cos∠FDE

∴ (EF)² = (6)² + (11)² - 2(6)(11) cos(40)

∴ (EF)² = 55.882133 ⇒ take square root for both sides

∴ EF = 7.47543 ≅ 7.48 feet

* The length of EF = 7.48 feet

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1 year ago

A standard weight known to weigh 10 grams. Some suspect bias in weights due to manufacturing process. To assess the accuracy of

notsponge [240]

Answer:

a) The 98% confidence interval for the mean weight is between 10.00409 grams and 10.00471 grams

b) 49 measurements are needed.

Step-by-step explanation:

Question a:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.98}{2} = 0.01$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.01 = 0.99$ , so Z = 2.327.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 2.327\frac{0.0003}{\sqrt{5}} = 0.00031$

The lower end of the interval is the sample mean subtracted by M. So it is 10.0044 - 0.00031 = 10.00409 grams

The upper end of the interval is the sample mean added to M. So it is 10 + 0.00031 = 10.00471 grams

The 98% confidence interval for the mean weight is between 10.00409 grams and 10.00471 grams.

(b) How many measurements must be averaged to get a margin of error of +/- 0.0001 with 98% confidence?

We have to find n for which M = 0.0001. So

$M = z\frac{\sigma}{\sqrt{n}}$

$0.0001 = 2.327\frac{0.0003}{\sqrt{n}}$

$0.0001\sqrt{n} = 2.327*0.0003$

$\sqrt{n} = \frac{2.327*0.0003}{0.0001}$

$(\sqrt{n})^2 = (\frac{2.327*0.0003}{0.0001})^2$

$n = 48.73$

Rounding up

49 measurements are needed.

7 0

1 year ago