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Alina [70]
3 years ago
8

The formula for the volume of a sphere is v= 4/3 m^3 what is the formula solved for r ?

Mathematics
1 answer:
Softa [21]3 years ago
4 0

Answer:

            3V

r = ∛ ( ---------- )

            4π

Step-by-step explanation:

Please, enclose the fraction 4/3 inside parentheses, to eliminate any possibility of misreading this fraction.  Also note that this formula MUST include "pi," symbolized by π.

V = (4/3) π r³  This formula does NOT include "m," which is a unit of measurement, not a variable.

Our task is to solve this formula for the radius, r.

Divide both sides by (4/3) π, to isolate r³.  This results in:

      v               (4/3) π r³

-------------   =  -----------------

  (4/3) π             (4/3) π

                     

                       V              3V

Then r³ = --------------  =  --------

                   (4/3) π           4π

and r is found by taking the cube root of the above result:

            3V

r = ∛ ( ---------- )

            4π

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Consider a t distribution with 7 degrees of freedom. Compute P(-1.29 < t < 1.29). Round your answer to at least three deci
ss7ja [257]

Answer:

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b) -1.73406361

Step-by-step explanation:

a)

Consider a t distribution with 7 degrees of freedom. Compute P(-1.29 < t < 1.29)

P(-1.29 < t < 1.29) would be the area under the t distribution curve with 7 degrees of freedom between -1.29 and 1.29, that is in the interval (-1.29, 1.29).

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So TDIST(1.29,7,2) gives the area outside (-1.29, 1.29).

If we subtract this value from 1 we get the desired result

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P(-1.29 < t < 1.29) = 1 - TDIST(1.29,7,2) = 1 - 0.23802914 = 0.76197086

In OpenOffice Calc, the function is the same replacing “,” with “;”  

That is

P(-1.29 < t < 1.29) = 1 - TDIST(1.29;7;2) = 0.76197086

b)

Consider a t distribution with 18 degrees of freedom. Find the value of c such that P(t≤ c) = 0.05

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TINV(p*2,n) with p>0 gives the point c such that the area of the t distribution with n degrees of freedom to the right of c is p.  

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So we want to compute  in Excel

-TINV(0.05*2,18) = -1.73406361

In OpenOffice Calc  

-TINV(0.05*2;18) = -1.73406361

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Answer:

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