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maxonik [38]
2 years ago
10

An investor has $ 500000 to spend. There investments are being considered, each having an expected annual interest rate. The int

erest rates are 15, 10 & 18 percent respectively. The investor`s goal is an average return of 15 percent in the three investments. Because of the high return on investment alternatives, the investor wants the amount in this alternative to equal 40 percent of the total investment. Determine whether there is a meaningful investment strategy which will satisfy these requirements. (BY Crammer Rule)
Mathematics
1 answer:
enyata [817]2 years ago
8 0
That's a lot of money and words. I don't appreciate this. I would probably just break down and cry tbh
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In a sale, normal prices are reduced by 20%. The sale price of a shirt is £26. Calculate the normal price of the shirt
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26/80*100 = <span>£</span>32.50 is your answer
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Help please!! 20 point question!! Which equation represents the general form a circle with a center at (-2, -3) and a diameter o
Ivan

Answer:

it is c

Step-by-step explanation:

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Please help me with this math question
earnstyle [38]
80.64
 because 16 times 12 is 192 
if every tile is 2 feet divide 192 by 2 
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6 0
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Natasha is seeding her backyard. The backyard is square in shape and has an area of 4225 square feet. What is the length of one
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2 years ago
Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its average number of unoc
Bingel [31]

Answer:

We need a sample size of at least 719

Step-by-step explanation:

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

\alpha = \frac{1-0.95}{2} = 0.025

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

So it is z with a pvalue of 1-0.025 = 0.975, so z = 1.96

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.

How large a sample size is required to vary population mean within 0.30 seat of the sample mean with 95% confidence interval?

This is at least n, in which n is found when M = 0.3, \sigma = 4.103. So

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

0.3 = 1.96*\frac{4.103}{\sqrt{n}}

0.3\sqrt{n} = 1.96*4.103

\sqrt{n} = \frac{1.96*4.103}{0.3}

(\sqrt{n})^{2} = (\frac{1.96*4.103}{0.3})^{2}

n = 718.57

Rouding up

We need a sample size of at least 719

6 0
2 years ago
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