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Dennis_Churaev [7]
3 years ago
9

Find the height of a cone with a volume of 150 in^3 and a diameter of 20 in. Use 3.14 for π.

Mathematics
1 answer:
Zolol [24]3 years ago
6 0

Volume = \frac{1}{3} \pi r^{2} h

150 = \frac{1}{3} x 3.14 x 10 x 10 x h

150 x 3 = 314 x h

h = 450/314

h = 1.43 inches

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Answer:

a) df=n-1=16-1=15

The statistic calculated is given by t=3.3  

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

p_v =P(t_{15}>3.3)=0.0024  

So since the p value is lower than the significance level pv we reject the null hypothesis.

b) df=n-1=8-1=7

The statistic calculated is given by t=1.8  

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

p_v =P(t_{7}>1.8)=0.057  

So since the p value is higher than the significance level pv>\alpha we FAIL to reject the null hypothesis.

c) df=n-1=26-1=25

The statistic calculated is given by t=-0.6  

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

p_v =P(t_{25}>-0.6)=0.723  

So since the p value is higher than the significance level pv>\alpha we FAIL to reject the null hypothesis.

Step-by-step explanation:

1) Data given and notation      

\bar X represent the sample mean

s represent the standard deviation for the sample

n sample size      

\mu_o =20 represent the value that we want to test    

\alpha 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)  

State the null and alternative hypotheses.      

We need to conduct a hypothesis in order to determine if the true mean is higher than 20, the system of hypothesis would be:      

Null hypothesis:\mu \leq 20      

Alternative hypothesis:\mu > 20      

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:      

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

(a) n = 16, t = 3.3, a = 0.05, P-value =

First we need to calculate the degrees of freedom given by:  

df=n-1=16-1=15

The statistic calculated is given by t=3.3  

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

p_v =P(t_{15}>3.3)=0.0024  

So since the p value is lower than the significance level pv we reject the null hypothesis.

(b) n = 8, t = 1.8, a = 0.01, P-value =

First we need to calculate the degrees of freedom given by:  

df=n-1=8-1=7

The statistic calculated is given by t=1.8  

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

p_v =P(t_{7}>1.8)=0.057  

So since the p value is higher than the significance level pv>\alpha we FAIL to reject the null hypothesis.

(c) n = 26,t = -0.6, P-value =

 First we need to calculate the degrees of freedom given by:  

df=n-1=26-1=25

The statistic calculated is given by t=-0.6  

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

p_v =P(t_{25}>-0.6)=0.723  

So since the p value is higher than the significance level pv>\alpha we FAIL to reject the null hypothesis.

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We'll use the Law of Cosines
We'll call
side a = length 5
side b = length 8
side c = length 10
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cos (C) = (5^2 + 8^2 - 10^2) / 2 * 5 * 8
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cos (C) = -11 / 80
cos (C) = <span> <span> <span> -0.1375 </span> </span> </span>
arc cosine ( <span> <span> <span> -0.1375 </span> </span> </span> ) = 97.903 Degrees
The largest angle is 97.903 Degrees



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3 years ago
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