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d1i1m1o1n [39]
2 years ago
11

A part of linear function g is graphed on the grid.

Mathematics
1 answer:
shutvik [7]2 years ago
5 0

By looking at the graph, we can see that:

  • Domain: -7 ≤ x ≤ 6
  • Range: -4 ≤ y ≤ 5.

<h3>How to find the domain and range of a function?</h3>

For a function:

y = f(x)

We define the domain as the set of the possible inputs x, and the range as the set of the possible outputs y.

Here, to find the domain, we need to see the range of "horizontal values" that the graph takes. We can see that it goes from -7 to 6, then the domain is:

-7 ≤ x ≤ 6

The range goes from -4 to 5, so the range is: -4 ≤ y ≤ 5.

Then the correct option is the first one.

If you want to learn more about domain and range, you can read:

brainly.com/question/1770447

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Suppose that 500 parts are tested in manufacturing and 10 are rejected.
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Answer:

a) z=\frac{0.02 -0.03}{\sqrt{\frac{0.03(1-0.03)}{500}}}=-1.31  

p_v =P(Z  

If we compare the p value obtained and using the significance level given \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 proportion of rejected items is less than 0.03.  

b) We can use a 95 percent upper confidence interval.

On this case we want a interval on this form : (-\infty,\hat p +z_{\alpha}\sqrt{\frac{\hat p (1-\hat p)}{n}})

So the critical value would be on this case Z_{\alpha}=1.64 and we can use the following excel code to find it: "=NORM.INV(1-0.05,0,1)"

We found the upper limit like this:

0.02+1.64\sqrt{\frac{0.02 (1-0.02)}{500}}=0.03026

And the interval would be: (-\infty,0.03026)

And since our value (0.02) is contained in the interval We fail to reject the hypothesis that p=0.03

Step-by-step explanation:

Part a

Data given and notation  

n=500 represent the random sample taken

X=10 represent the number of objects rejected

\hat p=\frac{10}{500}=0.02 estimated proportion of objects rejected

p_o=0.03 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)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that 70% of adults say that it is morally wrong to not report all income on tax returns.:  

Null hypothesis:p=0.03  

Alternative hypothesis:p < 0.03  

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{0.02 -0.03}{\sqrt{\frac{0.03(1-0.03)}{500}}}=-1.31  

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=0.05. The next step would be calculate the p value for this test.  

Since is a one tailed left test the p value would be:  

p_v =P(Z  

If we compare the p value obtained and using the significance level given \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 proportion of rejected items is less than 0.03.  

Part b

We can use a 95 percent upper confidence interval.

On this case we want a interval on this form : (-\infty,\hat p +z_{\alpha}\sqrt{\frac{\hat p (1-\hat p)}{n}})

So the critical value would be on this case Z_{\alpha}=1.64 and we can use the following excel code to find it: "=NORM.INV(1-0.05,0,1)"

We found the upper limit like this:

0.02+1.64\sqrt{\frac{0.02 (1-0.02)}{500}}=0.03026

And the interval would be: (-\infty,0.03026)

And since our value (0.02) is contained in the interval We fail to reject the hypothesis that p=0.03

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Step-by-step explanation:

idk if this is right but I tried

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