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Natali5045456 [20]
3 years ago
6

Come help a sister out on 2 geometry questions please

Mathematics
1 answer:
PSYCHO15rus [73]3 years ago
3 0
1- question

Data:
width = 6 cm
length = 5 cm
height = 8 cm


<span>First we will find the area of ​​the base that is square, we have:
</span>
Ab = 6 * 5 
Ab = 30 cm²

<span>Now we will calculate the volume of a square pyramid, we have:
</span>Formula: V =  \frac{1}{3}  A_{b}  *h

Solving:
V = \frac{1}{3} A_{b} *h
V =  \frac{1}{3}*30*8
V =  \frac{240}{3}
\boxed{\boxed{V = 80\:cm^3}}\end{array}}\qquad\quad\checkmark

Answer:
D) 80 cm³

2- question:

Data:
width = 7 in
length = 7 in
height = 9 in


First we will find the area of ​​the base that is square, we have:

Ab = 7 * 7 
Ab = 49 in²

Now we will calculate the volume of a square pyramid, we have:
Formula: V = \frac{1}{3} A_{b} *h

Solving:
V = \frac{1}{3} A_{b} *h
V = \frac{1}{3}*49*9
V = \frac{441}{3}
\boxed{\boxed{V = 147\:in^3}}\end{array}}\qquad\quad\checkmark

Answer: 
C) 147 in³

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Answer:

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Let Gabe = x

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Zappos is an online retailer based in Nevada and employs 1,300 employees. One of their competitors, Amazon, would like to test t
Amanda [17]

Answer:

t=\frac{33.9-36}{\frac{4.1}{\sqrt{22}}}=-2.402    

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t_{\alpha/2}= -2.08

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Step-by-step explanation:

Data given

\bar X=33.9 represent the sample mean

s=4.1 represent the sample standard deviation

n=22 sample size  

\mu_o =26 represent the value that we want to test

\alpha=0.025 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)  

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The statistic is given by:

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

And replacing we got:

t=\frac{33.9-36}{\frac{4.1}{\sqrt{22}}}=-2.402    

Now we can calculate the critical value but first we need to find the degreed of freedom:

df = n-1= 22-1=21

So we need to find a critical value in the t distribution with df =21 who accumulates 0.025 of the area in the left and we got:

t_{\alpha/2}= -2.08

Since the calculated values is lower than the critical value we have enough evidence to reject the null hypothesis at the significance level of 2.5% and we can say that the true mean is lower than 36 years old

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The numerical coefficient of this term is -3.
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