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2 years ago

If you have an 80 average so far in class after 3 tests what would you need to get on the fourth test to have an average of 83?

Mathematics

1 answer:

Monica [59]2 years ago

8 0

Answer:

Step-by-step explanation:

83*4 = 332

And 80*3 = 240

332-240 = 92

So you would need 92

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2 years ago

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1 year ago

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2 years ago

A bin is constructed from sheet metal with a square base and 4 equal rectangular sides. if the bin is constructed from 48 square

kondaur [170]

This is a problem of maxima and minima using derivative.

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V = length x width x height

That is:

$V = xxy = x^{2}y$

We also know that the <span>bin is constructed from 48 square feet of sheet metal, s</span>o:

Surface area of the square base = $x^{2}$

Surface area of the rectangular sides = $4xy$

Therefore, the total area of the cube is:

$A = 48 ft^{2} = x^{2} + 4xy$

Isolating the variable y in terms of x:

$y = \frac{48- x^{2} }{4x}$

Substituting this value in V:

$V = x^{2}( \frac{48- x^{2} }{x}) = 48x- x^{3}$

Getting the derivative and finding the maxima. This happens when the derivative is equal to zero:

$\frac{dv}{dx} = 48-3x^{2} =0$

Solving for x:

$x = \sqrt{\frac{48}{3}} = \sqrt{16} = 4$

Solving for y:

$y = \frac{48- 4^{2} }{(4)(4)} = 2$

Then, <span>the dimensions of the largest volume of such a bin is:
</span>
Length = 4 ft
Width = 4 ft
Height = 2 ft

And its volume is:

$V = (4^{2} )(2) = 32 ft^{3}$

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2 years ago

If you have an 80 average so far in class after 3 tests what would you need to get on the fourth test to have an average of 83?​

If you have an 80 average so far in class after 3 tests what would you need to get on the fourth test to have an average of 83?