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Anvisha [2.4K]
3 years ago
7

Deon invested his savings in two investment funds. The amount he invested in Fund A was 8000 less than the amount he invested in

Fund B. Fund A returned a 5% profit and Fund B returned a 7% profit. How much did he invest in Fund B, if the total profit from the two funds together was 1760?
Mathematics
1 answer:
luda_lava [24]3 years ago
7 0

9514 1404 393

Answer:

  • A: 10,000
  • B: 18,000

Step-by-step explanation:

Let A represent the amount invested in fund A, Then A+8000 is the amount invested in fund B, and the total return is ...

  0.05A +0.07(A+8000) = 1760

  0.12A +560 = 1760

  0.12A = 1200

  A = 10000

Deon invested 10,000 in fund A and 18,000 in fund B.

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In doing so, you collect a random sample of 50 salespersons employed by his company, which is thought to be representative of sa
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Answer:

z=\frac{0.36 -0.45}{\sqrt{\frac{0.45(1-0.45)}{50}}}=-1.279  

p_v =P(z

And we can use the following code to find it  "=NORM.DIST(-1.279,0,1,TRUE)"

Step-by-step explanation:

Assuming this complete problem: "The CEO of a software company is committed to expanding the proportion of highly qualified women in the organization's staff of salespersons. He believes that the proportion of women in similar sales positions across the country is less than 45%. Hoping to find support for his belief, he directs you to test

H0: p .45 vs H1: p < .45.

In doing so, you collect a random sample of 50 salespersons employed by his company, which is thought to be representative of sales staffs of competing organizations in the industry. The collected random sample of size 50 showed that only 18 were women.

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1) Data given and notation

n=50 represent the random sample taken

X=18 represent the number of women in the sample selected

\hat p=\frac{18}{50}=0.36 estimated proportion of women in the sample

p_o=0.45 is the value that we want to test

\alpha represent the significance level

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion of women is less than 0.45:  

Null hypothesis:p\geq 0.45  

Alternative hypothesis:p < 0.45  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.36 -0.45}{\sqrt{\frac{0.45(1-0.45)}{50}}}=-1.279  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

p_v =P(z

And we can use the following code to find it  "=NORM.DIST(-1.279,0,1,TRUE)"

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Answer:

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As per the statement:

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