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jenyasd209 [6]
3 years ago
15

-3x + 2y = 7 What is the answer to this algebra question

Mathematics
1 answer:
Salsk061 [2.6K]3 years ago
7 0
Answer: 0
First: Move constant to the left by adding its opposite to both sides
-3x+2y-7=7-7
Then: Change the signs on both sides of the equation
3x-2y+7
Equaling: 3x-2y+7=0
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timofeeve [1]

Answer:

Step-by-step explanation:

not sure srorry for replying dont look at this

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What is the solution of the following system of linear equations? −21 + 15x = 6y −5x + 2y = −8 Select one: A. (15, -7) B. (4, 11
quester [9]

Answer:

C. No solution

Step-by-step explanation:

System of Equations:

A) -21+15x=6y

B) -5x+2y=-8

Simplifying and rearranging equation A.

Dividing each term by 3 (common factor for each term) in equation A.

\frac{-21}{3}+\frac{15x}{3}=\frac{6y}{2}

-7+5x=2y

Subtracting both sides by 2y

-7+5x-2y=2y-2y

-7+5x-2y=0

Adding 7 both sides.

-7+7+5x-2y=0+7

5x-2y=7  

Adding the above equation to equation B.

   5x-2y=7  

+ -5x+2y=-8

We have 0=-1

which is not true.

Hence the system has no solution. (Answer)

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3 years ago
A train is traveling 300 feet per second. How fast is this in miles per hour?
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Answer:

204.545 miles per hour

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A study by the Pew Research Center1 reports that in 2010, for the first time, more adults aged 18 to 29 got their news from the
sergejj [24]

Answer:

Null hypothesis:p \leq 0.65  

Alternative hypothesis:p > 0.65  

z=\frac{0.66 -0.65}{\sqrt{\frac{0.65(1-0.65)}{1500}}}=0.812  

p_v =2*P(z>0.812)=0.208  

So the p value obtained was a very high value and using the significance level assumed \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the true proportion it's not significantly higher from 0.65.  

We don't have enough evidence to conclude that the true % it's higher than 65%, since we fail to reject the null hypothesis on this case.

Step-by-step explanation:

1) Data given and notation

n=1500 represent the random sample taken

X represent the adults that   said television was one of their main sources of news.

\hat p=0.66 estimated proportion of adults that said television was one of their main sources of news.

p_o=0.65 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

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 the proportion is higher than 0.65:  

Null hypothesis:p \leq 0.65  

Alternative hypothesis:p > 0.65  

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{0.66 -0.65}{\sqrt{\frac{0.65(1-0.65)}{1500}}}=0.812  

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 right tailed test the p value would be:  

p_v =P(z>0.812)=0.208  

So the p value obtained was a very high value and using the significance level assumed \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the true proportion it's not significantly higher from 0.65.  

Do we find evidence that more than 65% of all US adults use television as one of their main sources for news?

We don't have enough evidence to conclude that the true % it's higher than 65%, since we fail to reject the null hypothesis on this case.

6 0
3 years ago
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