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3 years ago

John got a 76, 68, and 83 on his first three test. What score does he need on his fourth test to achieve an 80 average?

1 answer:

3 0

It's 93.
simple calculate it by:
(76+68+83+x)/4=80
then 73*5-(67+75+73+69)=x
where x here is the grade needed to reach an av. of 80. 4 is the number of tests.
*common answer*

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If the measure of the given central angle is $\theta$ rad, then the length of the subtended arc is $\ell$ satisfying

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Please solve with explanation I’ve been asking all day (this is not a multiple choice question)

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a) $SA = 522.9~cm^2$

b) $V_{cone} = 670.2~cm^3$

c) $V_{empty} = 1340.4~cm^3$

Step-by-step explanation:

For a cone,

$SA = \pi r (L + r)$

where L = slant height

$L = \sqrt{r^2 + h^2}$

We have r = 8 cm; h = 10 cm

$L = \sqrt{(8~cm)^2 + (10~cm)^2}$

$L = \sqrt{164~cm^2}$

$SA = (\pi)(8~cm)(\sqrt{164~cm^2} + 8~cm)$

$SA = 522.9~cm^2$

$V_{cone} = \dfrac{1}{3}\pi r^2 h$

$V_{cone} = \dfrac{1}{3}(\pi)(8~cm)^2(10~cm)$

$V_{cone} = 670.2~cm^3$

$V_{cylinder} = \pi r^2 h$

empty space = volume of cylinder - volume of cone

$V_{empty} = V_{cylinder} - V_{cone}$

$V_{empty} = \pi r^2 h - \dfrac{1}{3}\pi r^2 h$

$V_{empty} = (\pi)(8~cm)^2(10~cm) - \dfrac{1}{3}(\pi)(8~cm)^2(10~cm)$

$V_{empty} = 1340.4~cm^3$

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