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Jlenok [28]
3 years ago
9

The work of a student to solve a set of equations is shown below:

Mathematics
2 answers:
Zolol [24]3 years ago
3 0
He made erron in step 3. He forgot add 4. it should be  
-4m + 4m = -32 - 8n + 4n +4

Greeting
marusya05 [52]3 years ago
3 0
The error is found in step 3.

Here is what he did:

-4m + 4m = -32 - 8n + 4n <<<<<<<<<<<< What happened to +4???

0 = -32 - 4n

n = -8.

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Solve this equation 3+5/2x=1/2x-8
guajiro [1.7K]

Answer:

34

Step-by-step explanation:

3 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
What is f(g(x)) for x &gt; 5?
mel-nik [20]

Answer:

\large\boxed{B.\ 4x^2-41x+105}

Step-by-step explanation:

f(x)=4x-\sqrt{x}\\\\g(x)=(x-5)^2\\\\f(g(x))\to\text{put}\ x=(x-5)^2\ \text{to}\ f(x):\\\\f(g(x))=f\bigg((x-5)^2\bigg)=4(x-5)^2-\sqrt{(x-5)^2}\\\\\text{use}\\(a-b)^2=a^2-2ab+b^2\\\sqrt{x^2}=|x|\\\\f(g(x)=4(x^2-2(x)(5)+5^2)-|x-5|\\\\x>5,\ \text{therefore}\ x-5>0\to|x-5|=x-5\\\\f(g(x))=4(x^2-10x+25)-(x-5)\\\\\text{use the distributive property:}\ a(b+c)=ab+ac\\\\f(g(x))=(4)(x^2)+(4)(-10x)+(4)(25)-x-(-5)\\\\f(g(x))=4x^2-40x+100-x+5\\\\\text{combine like terms}\\\\f(g(x))=4x^2+(-40x-x)+(100+5)\\\\f(g(x))=4x^2-41x+105

6 0
3 years ago
Read 2 more answers
how many square feet will cover a rectangle 60 feet by 20 feet? Explain how you determined your answer
nika2105 [10]

Answer:

1200

Step-by-step explanation:

2 times 6 = 12

12 times 100 = 1200

Hope this helps if does not please let me know thanks!

7 0
3 years ago
Pls help is due tomorrow and if u help i will give u brainliest
Reil [10]

Answer:

Search it up

Step-by-step equation:

Just search up the website at the bottom of the paper.

6 0
3 years ago
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