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Svetach [21]
3 years ago
6

A combined total of 39000 is invested in two bonds that pay 5% and 6.5% simple interest. The annual interest is 235500 how much

is invested in each bond?
Mathematics
1 answer:
dolphi86 [110]3 years ago
4 0

Answer:

$12000 at 5%

$27000 at 6.5%.

Step-by-step explanation:

Let x represent amount invested at 5% and y represent amount invested at 6.5%.

We have been given that a combined total of 39000 is invested in two bonds. We can represent this information in an equation as:

x+y=39000...(1)

We are also told that the annual interest rate is 2355.00. We can represent this information in an equation as:

0.05x+0.065y=2355...(2)

Upon substituting equation (1) in equation (2), we will get:

0.05x+0.065(39000-x)=2355

0.05x+2535-0.065x=2355

-0.015x+2535=2355

-0.015x=2355-2535

-0.015x=-180

\frac{-0.015x}{-0.015}=\frac{-180}{-0.015}

x=12000

Therefore, an amount of $12,000 is invested at 5%.

Upon substituting x=12000 in equation (1), we will get:

12000+y=39000

y=39000-12000

y=27000

Therefore, an amount of $27,000 is invested at 6.5%.

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The last thing to do is to subtract the areas of the two circles and blue rectangle from the area of the entire thing:

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Null hypothesis:p \leq 0.69  

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Step-by-step explanation:

1) Data given and notation

n=1500 represent the random sample taken

X represent the number of graduates that had student loan debt in 2014

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p_v represent the p value (variable of interest)  

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We need to conduct a hypothesis in order to test the claim that if there was a significant increase in the proportion of student loan debt for public and nonprofit colleges in 2014 respect to the value of 2013.:  

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

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