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andrew11 [14]
3 years ago
8

2. The total weight of 4 tires is 800 pounds. What is the unit weight of 1 tire?

Mathematics
2 answers:
Artemon [7]3 years ago
7 0

Answer:

The weight unit is 200 pounds per tire

Kazeer [188]3 years ago
3 0
The tire is 200 pounds for every tire
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The scale factor for two similar rectangles is 4. If the area of the smaller rectangle 7m^2, what is the area of the larger rect
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Answer:

A. 112 m^2

Step-by-step explanation:

If the scale factor is 4, the area of the larger figure is 4^2 times that of the smaller one.

... (7 m^2)·4^2 = 112 m^2

_____

The scale factor for area is the square of the scale factor for linear dimensions. When we talk about a scale factor without qualification, we mean the scale factor for linear dimensions.

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An NBA fan named Mark claims that there are more fouls called on his team 1 point
inn [45]

Answer:

t=\frac{12.2-11.5}{\frac{1.6}{\sqrt{34}}}=2.551    

df = n-1=34-1=33

p_v =P(t_{(33)}>2.551)=0.008  

Since the p value is less than the significance level of 0.05 we have enough evidence to reject the null hypothesis in favor of the claim

And the best conclusion for this case would be:

b)The p-value is 0.008, indicating sufficient evidence for his claim.

Step-by-step explanation:

Information provided

\bar X=12.2 represent the sample mean fould against

s=1.6 represent the sample standard deviation

n=34 sample size  

represent the value that we want to test

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

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

System of hypothesis

We need to conduct a hypothesis in order to check if the true mean is higher than 11.5 fouls per game:  

Null hypothesis:\mu \leq 11.5  

Alternative hypothesis:\mu > 11.5  

The statistic is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

The statistic is given by:

t=\frac{12.2-11.5}{\frac{1.6}{\sqrt{34}}}=2.551    

P value

The degreed of freedom are given by:

df = n-1=34-1=33

Since is a one side test the p value would be:  

p_v =P(t_{(33)}>2.551)=0.008  

Since the p value is less than the significance level of 0.05 we have enough evidence to reject the null hypothesis in favor of the claim

And the best conclusion for this case would be:

b)The p-value is 0.008, indicating sufficient evidence for his claim.

7 0
3 years ago
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