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Nata [24]
4 years ago
5

What does it mean to evaluate a function?

Mathematics
2 answers:
mario62 [17]4 years ago
7 0
To solve the function

netineya [11]4 years ago
4 0
What is means is the...

Uhh...... I actually don't know, I never thought of that actually...
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Solve for x.
andre [41]

Answer:

A cuz x+6 is double x-3 so divide them and equate it to 2 you'd get x+6 = 2x-6

12=2x-x

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3 years ago
If m<1 is 115 what is m<4​
aniked [119]

I'm not sure if this is right, but I multiplied 115 by 4 and got 460- Try that? ;^;

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3 years ago
A research center claims that at least 4545​% of adults in a certain country think the government is not aggressive enough in pu
frez [133]

Answer:

z=\frac{0.39 -0.45}{\sqrt{\frac{0.45(1-0.45)}{700}}}=-3.19  

p_v =P(z  

So the p value obtained was a very low value and using the significance level given \alpha=0.01 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of interst is significantly lower than 0.45 and the claim is not appropiate

Step-by-step explanation:

Data given and notation

n=700 represent the random sample taken

\hat p=0.39 estimated proportion of interest

p_o=0.7 is the value that we want to test

\alpha=0.01 represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is at least 0.45.:  

Null hypothesis:p\geq 0.45  

Alternative hypothesis:p < 0.45  

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.

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.39 -0.45}{\sqrt{\frac{0.45(1-0.45)}{700}}}=-3.19  

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 provided \alpha=0.01. The next step would be calculate the p value for this test.  

Since is a lef tailed test the p value would be:  

p_v =P(z  

So the p value obtained was a very low value and using the significance level given \alpha=0.01 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of interst is significantly lower than 0.45 and the claim is not appropiate

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