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lisov135 [29]
3 years ago
14

Edwin deposited money into a savings account that pays a simple annual interest rate of 1.4%. He

Mathematics
1 answer:
mafiozo [28]3 years ago
5 0
<h3>The deposited sum of money by Edwin is $350.</h3>

Step-by-step explanation:

Let us assume that the amount deposited  = P

The rate of interest  = 1.4%

The Simple Interest earned  = $24.50

Time  = T = 5 years

Now, as we know:

SIMPLE INTEREST  = \frac{P \times R \times T}{100}

\implies 24.50  = \frac{P \times 1.4 \times  5}{100} \\\implies P = \frac{24.50 \times 100}{1.4 \times 5}  = 350\\\implies P = 350

Hence, the deposited sum of money by Edwin is $350.

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HELP QUICK LIKE RIGHT NOW!please
pogonyaev

Answer:

1/2 = 1/2

Step-by-step explanation:

5/10 is equal to 1/2 and 1/2 cant get smaller making you answer 1/2 = 1/2

8 0
3 years ago
Read 2 more answers
Does going to a private university increase the chance that a student will graduate with student loan debt? A national poll by t
Snowcat [4.5K]

Answer:

Null hypothesis:p \leq 0.69  

Alternative hypothesis:p > 0.69  

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

\hat p=0.71 estimated proportion of adults that said that it is morally wrong to not report all income on tax returns

p_o=0.69 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 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.:  

Null hypothesis:p \leq 0.69  

Alternative hypothesis:p > 0.69  

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{0.71 -0.69}{\sqrt{\frac{0.69(1-0.69)}{1500}}}=1.675  

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 is \alpha=0.05. The next step would be calculate the p value for this test.  

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

p_v =P(Z>1.675)=0.047  

If we compare the p value obtained 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, and we can said that at 5% of significance the proportion of adults that said that it is morally wrong to not report all income on tax returns  is not significantly higher than 0.69.  

5 0
3 years ago
Rename 28 percent as a decimal and fraction in its simplest form.
VARVARA [1.3K]
28% is 28/100 as a fraction but the simplest form is 14/50 and it can be simplified to 7/25. the decimal is 0.28

fraction is 7/25
decimal is 0.28

7 0
3 years ago
Read 2 more answers
I need to simplify this
Anon25 [30]
Your answer would be
-7-(-i)+6-3i
3 0
3 years ago
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Solve for C.<br> -5/8c=20
iris [78.8K]
-5/8c = 20....multiply both sides by -8/5 (this cancels out the -5/8 on the left
c = 20(-8/5)
c = -160/5
c = - 32

check..
-5/8(-32) = 20
160/8 = 20
20 = 20 (correct)

so c = -32
8 0
3 years ago
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