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Vika [28.1K]
3 years ago
7

This is similar to the last one but this is also an quiz that’s worth a major. Can anybody help?

Mathematics
1 answer:
VashaNatasha [74]3 years ago
3 0

Answer:

42

Step-by-step explanation:

Shanice: 88

Greg: (88 - 4) ÷ 2 = 84 ÷ 2 = 42

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Need the missing length
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10

Step-by-step explanation:

The side to the right is 15. The side to the left is 5. You can subtract these two to find the length.

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Terry is 1.78 m tall.<br> Steven is 9cm taller then Terry.<br> How tall is Steven?
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11. The perimeter of a rectangular garden is 96 ft. The length is five times the width. Find
asambeis [7]

Answer:

Length =40 ft.

width. = 8 ft.

Step-by-step explanation:

If The length is five times the width , then

width = w

length = 5w

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2( 6w) = 96

12 w= 96

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= 40 ft

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3 years ago
If the Minimum is 68, and the Maximum is 97, what is the Range of the Box-and-Whisker Plot?
spayn [35]

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b

Step-by-step explanation:

6 0
3 years ago
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Ten years ago 53% of American families owned stocks or stock funds. Sample data collected by the Investment Company Institute in
Alborosie

Answer:

a) Null hypothesis:p\geq 0.53  

Alternative hypothesis:p < 0.53  

b) z=\frac{0.46 -0.53}{\sqrt{\frac{0.53(1-0.53)}{300}}}=-2.429  

p_v =P(Z

c) 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 American families owning stocks or stock funds is significantly less than 0.53 .  

Step-by-step explanation:

Data given and notation

n=300 represent the random sample taken

\hat p=0.46 estimated proportion of American families owning stocks or stock funds

p_o=0.53 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  

Part a

We need to conduct a hypothesis in order to test the claim that proportion is less than 0.53 or 53%.:  

Null hypothesis:p\geq 0.53  

Alternative hypothesis:p < 0.53  

Part b

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{0.46 -0.53}{\sqrt{\frac{0.53(1-0.53)}{300}}}=-2.429  

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 left tailed test the p value would be:  

p_v =P(Z

Part c  

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 American families owning stocks or stock funds is significantly less than 0.53 .  

7 0
3 years ago
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