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cricket20 [7]
3 years ago
11

A retired woman has $280,000 to invest. She has chosen one relatively safe investment fund that has an annual yield of 9% and an

other riskier fund that has a 13% annual yield. How much should she invest in each fund if she would like to earn $28,000 per year from her investments?
9% fund $

13% fund $
Mathematics
1 answer:
ASHA 777 [7]3 years ago
6 0

Answer:

9% fund: $ 210,000

13% fund: $70,000

Step-by-step explanation:

As she wants to have a $28,000 annual return for her $280,000 investment, she is expecting a return rate of 10%:

r=\dfrac{R}{C}=\dfrac{28,000}{280,000}=0.10

If we call x the proportion of the capital in the 9% fund, then (1-x) is the proportion of the capital in the 13% fund,and the return of the combination has to be the expected return of 10%:

0.09x+0.13(1-x)=0.10\\\\0.09x+0.13-0.13x=0.10\\\\-0.04x=0.10-0.13=-0.03\\\\x=\dfrac{0.03}{0.04}=0.75

Then, we know that 75% of the capital should be invested in the 9% fund and 25% in the 13% fund.

This correspond to a capital of:

9% fund: 0.75*$280,000 = $ 210,000

13% fund: 0.25*$280,000 = $70,000

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1. last choice

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Step-by-step explanation:

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Consider a value to be significantly low if its z score less than or equal to minus−2 or consider a value to be significantly hi
ziro4ka [17]

Answer:

If the weight is higher than 5.8886 gr would be considered significantly high

If the weight is lower than 5.6121 gr would be considered significantly low

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

X \sim N(5.75241,0.06281)  

Where \mu=5.75241 and \sigma=0.06281

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

For the case when z =-2 we can do this:

-2 = \frac{X-5.75241}{0.06281}

And if we solve for X we got:

X = 5.75241 -2*0.06281 =5.6121

And for the other case when Z=2 we have:

2 = \frac{X-5.75241}{0.06281}

And if we solve for X we got:

X = 5.75241 +2*0.06281 =5.8886

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If the weight is lower than 5.6121 gr would be considered significantly low

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