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Neporo4naja [7]
3 years ago
12

PLEASE HELP ME!!!

Mathematics
2 answers:
zalisa [80]3 years ago
8 0
Answer:Is D

Explanation:i got it rigth
Irina-Kira [14]3 years ago
3 0

Answer:

size comes after color

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Solve the problem. Find the amount of money in an account after 12 years if $4700 is deposited at 5% annual interest compounded
Dmitrij [34]

Answer:

The correct answer is $8532.17

Step-by-step explanation:

The formula for calculating investments with compound interests is as follows:

(1+\frac{R}{t})^{tn}*P

Where:

R is the annual interest rate,

t is the number of times the investment is to be compounded in a year,

n is the number of years,

P is the principal amount invested.

Replacing in the formula with the given values you have:

(1+\frac{0.05}{4})^{4*12}*4700 = 8532.1678

3 0
3 years ago
Please help me on these two questions
irina [24]

Answer:A for the first one and C for the second one

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Please Help Explain for brainliest
Anettt [7]
I would say Song 1 on the top could be NO, and Song 1 on the bottom could be ON, then Song 2 on the very top could be NO again, then on the bottom of the very top could OO, then on the bottom of that could be NN, and on he very bottom could be ON.
Hope this helps.
3 0
3 years ago
How can I explain how to make a 3.5 into a fraction for Mt 10yr old?
stealth61 [152]
The number on the left side of the decimal, write it as a large, in this case, three. The number on the right side of the decimal, write as a small number. Then put a ten under the five, and divide the two with a line. A way to remember this is - Imagine you have 3 cookies. The .5 represents a half of a cookie. You need to have all ten parts of the cookie to make a whole cookie. You'd need 5 more pieces of cookie to make 4 cookies. Hope this helps.
8 0
3 years ago
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
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