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vekshin1
3 years ago
12

QUESTION 12 PLZ HELP ASAP

Mathematics
1 answer:
igor_vitrenko [27]3 years ago
6 0
12.
12.5 - 10 - 7.5  <  5.11 >
The answer would most likely be 5.11 or 11.5
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Write the hourly wage 15 dollars an hour as a fraction
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Answer:

x=-2.5,9

Step-by-step explanation:

Set them equal to each other to say 2x^2-9x=4x+45. Get them all on one side to get 2x^2-13x-45. Factor that out to get (x+2.5)(x-9). Set that equal to 0 to get x=-2.5,9

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Please help with math show your work
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Two different types of polishing solutions are being evaluated for possible use in a tumble-polish operation for manufacturing i
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Answer:

z=\frac{0.843-0.653}{\sqrt{0.748(1-0.748)(\frac{1}{300}+\frac{1}{300})}}=5.358    

p_v =2*P(Z>5.358) = 4.2x10^{-8}  

Comparing the p value with the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that we have singificantly differences between the two proportions.  

Step-by-step explanation:

Data given and notation  

X_{1}=253 represent the number with no defects in sample 1

X_{2}=196 represent the number with no defects in sample 1

n_{1}=300 sample 1

n_{2}=300 sample 2

p_{1}=\frac{253}{300}=0.843 represent the proportion of number with no defects in sample 1

p_{2}=\frac{196}{300}=0.653 represent the proportion of number with no defects in sample 2

z would represent the statistic (variable of interest)  

p_v represent the value for the test (variable of interest)  

\alpha=0.05 significance level given

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if is there is a difference in the the two proportions, the system of hypothesis would be:  

Null hypothesis:p_{1} - p_2}=0  

Alternative hypothesis:p_{1} - p_{2} \neq 0  

We need to apply a z test to compare proportions, and the statistic is given by:  

z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}   (1)  

Where \hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{253+196}{300+300}=0.748  

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.  

Calculate the statistic  

Replacing in formula (1) the values obtained we got this:  

z=\frac{0.843-0.653}{\sqrt{0.748(1-0.748)(\frac{1}{300}+\frac{1}{300})}}=5.358    

Statistical decision

Since is a two sided test the p value would be:  

p_v =2*P(Z>5.358) = 4.2x10^{-8}  

Comparing the p value with the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that we have singificantly differences between the two proportions.  

5 0
3 years ago
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