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quester [9]
3 years ago
13

What is a name of the line? A. I B. KJ C. J D. JI

Mathematics
2 answers:
Rudiy273 years ago
5 0

Answer:

Line JI (Option D)

Step-by-step explanation:

Always take the 2 points of the line to name the line.

zubka84 [21]3 years ago
3 0

Answer:

  D.  JI

Step-by-step explanation:

The line can be named any of ...

  • k . . .  name shown next to the line
  • JI . . . two points on the line
  • IJ . . . two points on the line

Of these, JI is on your choices list.

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Answer:

the answer would be 4.2

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The sides of a rectangle are in the ratio of 4:5. If the width is 28 in., find the length, the perimeter, and the area of this r
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If the width is 28 inches, then divide that by 4 and you get 7. You multiply that by 5 to get the length. That would be 35. Just to check, you know that the width 28 and length 35 are in ratio 4:5 if you divide by 7. The perimeter would be 2(35+28)=63*2=126. So the perimeter is 126. The area would be 35*28 which is 980. To sum up, the answers are as follows.

Length: 35 in
Perimeter: 126 in
Area: 980 inches squared.
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A cross-country skier traveled at a rate of 40 miles per hour on downhills and 20 miles per hour on flat terrain while skiing a
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A snail travels at a rate of 2.35 feet per minute. Write a rule to describe the function. How far will the snail travel in 5 min
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3 years ago
Suppose that 500 parts are tested in manufacturing and 10 are rejected.
alexdok [17]

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

3 0
3 years ago
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