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Jlenok [28]
2 years ago
10

A bank loaned out ​$19,000​, part of it at the rate of 9 % per year and the rest at 19 % per year. If the interest received in o

ne year totaled ​$2500​, how much was loaned at 9 % question mark
How much of the ​$19,000 did the bank loan out at 9 %?
Mathematics
1 answer:
VMariaS [17]2 years ago
7 0

Answer: $11100 was loaned at 9%

Step-by-step explanation:

Let x represent the amount that the

bank loaned out at the annual rate of 9%.

Let y represent the amount that the

bank loaned out at the annual rate of 19%.

The bank loaned out ​$19,000​, part of it at the rate of 9% per year and the rest at 19% per year. It means that

x + y = 19000

The interest on $x after 1 year would be

9/100 × x = 0.09x

The interest on $y after 1 year would be

19/100 × y = 0.19y

If the interest received in one year totaled ​$2500. It means that

0.09x + 0.19y = 2500- - - - - - - - - 1

Substituting x = 19000 - y into equation 1, it becomes

0.09(19000 - y) + 0.19y = 2500

1710 - 19000y + 0.19y = 2500

- 0.09y + 0.19y = 2500 - 1710

- 0.1y = - 790

y = - 790/- 0.1

y = $7900

x = 19000 - y = 19000 - 7900

x = $11100

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Step-by-step explanation:

1) Data given and notation  

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X=97 represent the students who completed the modules before class

\hat p=\frac{97}{110}=0.882 estimated proportion of students who completed the modules before class

p_o=0.8 is the value that we want to test

\alpha represent the significance level

z would represent the statistic (variable of interest)

p_v{/tex} 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 true proportion is less than 0.8 or 80%:  Null hypothesis:[tex]p\geq 0.8  

Alternative hypothesis:p < 0.8  

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{0.881 -0.8}{\sqrt{\frac{0.8(1-0.8)}{110}}}=2.124  

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 significance level assumed \alpha=0.05. 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  

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Alternative hypothesis:p > 0.8

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So the p value obtained was a very low value and using the significance level assumed \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis.

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