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melamori03 [73]
3 years ago
15

1/3(4y-2)+ 1/9(6y+10)

Mathematics
2 answers:
Soloha48 [4]3 years ago
6 0
Here are the steps :
1- Distributing

1/3 (4y-2) + 1/9 (6y + 10 )

1/3 ( 4y ) = 4/3 = 1 1/3 y
1/3 ( -2 ) = -2/3

1/9 ( 6y ) = 6/9 y
1/9 ( 10 ) = 10/9 or 1 1/9

Now we get the equation :
1 1/3y - 2/3 + 6/9y + 1 1/9


2- finding common denominators
1 1/3y and 6/9y can both have 9 as the denominator .
The fraction 1/3 has to be multiplied by 3 , both numerator & denominator , to get the same denominator (sorry if it’s confusing) .

Then we’d get 1 3/9y + 6/9y = 2y

Now we do the same thing with the other digits (without a variable)

- 2/3 + 1 1/9 or
- 6/9 + 1 1/9 = 4/9

3 - answer
So ,

1/3 (4y-2) + 1/9 (6y + 10 )

Distributed is

2y - 6/9



alukav5142 [94]3 years ago
3 0
The answers is 2y+4/9

Hope that helps!! :)
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When 3010 adults were surveyed in a​ poll, 27​% said that they use the Internet. Is it okay for a newspaper reporter to write th
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Answer:

Null hypothesis:p=0.25  

Alternative hypothesis:p \neq 0.25

z=2.53

pv=0.0114

So based on the p value obtained and using the significance given \alpha=0.05 we have p_v so we can conclude that we reject the null hypothesis, and we can said that at 5% of significance the proportion of people who says that they use the Internet differs from 0.25 or 25% .  

Step-by-step explanation:

1) Data given and notation

n=3010 represent the random sample taken

X represent the people who says that said that they use the Internet.

\hat p=\frac{X}{106}=0.27 estimated proportion of people who says that said that they use the Internet.

p_o=0.25 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 50% of people who says that  they would watch one of the television shows.:  

Null hypothesis:p=0.25  

Alternative hypothesis:p \neq 0.25  

When we conduct a proportion test we need to use the z statisitc, 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{0.27 -0.25}{\sqrt{\frac{0.25(1-0.25)}{3010}}}=2.53

4) Statistical decision  

P value method or p value approach . "This method consists on 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.  

We have the significance level provided \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

p_v =2*P(z>2.53)=2*(0.0057)=0.0114  

So based on the p value obtained and using the significance given \alpha=0.05 we have p_v so we can conclude that we reject the null hypothesis, and we can said that at 5% of significance the proportion of people who says that they use the Internet differs from 0.25 or 25% .  

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