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kozerog [31]
3 years ago
6

V=pir^2h solve this equation for h

Mathematics
2 answers:
murzikaleks [220]3 years ago
5 0
Hey!

First, let's write the problem.
V=\pi r^2h
To make it easier, let's switch sides.
\pi r^2h=V
Divide both sides by \pi r^2
\frac{\pi r^2h}{\pi r^2}=\frac{V}{\pi r^2}

Our final answer would be,
h=\frac{V}{\pi r^2}

Thanks!
-TetraFish
Nady [450]3 years ago
3 0
Since we are solving for h, the equation:

V = πr²h

...divides πr² to the other side. Thus, it becomes....

\frac{V}{ \pi r^2} = h

Which can be rewritten as....

h = \frac{V}{ \pi r^2}
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A lathe is set to cut bars of steel into lengths of 6 cm. The lathe is considered to be in perfect adjustment if the average len
Gnom [1K]

Answer:

t=\frac{5.97-6}{\frac{0.4}{\sqrt{93}}}=-0.723    

E. -0.723

df=n-1=93-1=92  

p_v =2*P(t_{(92)}  

Since the p value is very high we don't have enough evidence to conclude that the true mean for the lengths is different from 6 cm.

Step-by-step explanation:

Information provided

\bar X=5.97 represent the sample mean for the length

s=0.4 represent the sample standard deviation

n=93 sample size  

\mu_o =6 represent the value that we want to test

\alpha=0.05 represent the significance level

t would represent the statistic  

p_v represent the p value for the test

System of hypothesis

We need to conduct a hypothesis in order to check if the lathe is in perfect adjustment (6cm), then the system of hypothesis would be:  

Null hypothesis:\mu = 6  

Alternative hypothesis:\mu \neq 6  

since we don't know the population deviation the statistic is:

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

Replacing in formula (1) we got:

t=\frac{5.97-6}{\frac{0.4}{\sqrt{93}}}=-0.723    

E. -0.723

P value

The degrees of freedom are given by:

df=n-1=93-1=92  

Since is a two tailed test the p value would be:  

p_v =2*P(t_{(92)}  

Since the p value is very high we don't have enough evidence to conclude that the true mean for the lengths is different from 6 cm.

5 0
3 years ago
Simplify the given expression below 4/3-2i
ANTONII [103]
So-called simplifying, really means, "rationalizing the denominator", which is another way of saying, "getting rid of that pesky radical in the bottom"


\bf \cfrac{4}{3-2i}\cdot \cfrac{3+2i}{3+2i}\impliedby \textit{multiplying by the conjugate of the bottom}
\\\\\\
\cfrac{4(3+2i)}{(3-2i)(3+2i)}\implies \cfrac{4(3+2i)}{3^2-(2i)^2}\implies \cfrac{4(3+2i)}{3^2-(4i^2)}\\\\
-------------------------------\\\\
recall\qquad i^2=-1\\\\
-------------------------------\\\\
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Answer:

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Step-by-step explanation:

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Answer:

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g(x)=16 means that

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Q3

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