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antoniya [11.8K]

2 years ago

6

Jessica wants to calculate her grade in her accounting class.she has four test grades (82, 65, 71, and 77) and a homework grade

(92). her highest test grade counts for 25% of her grade, her lowest test grade counts for 15% of her grade, the two remaining tests count for 20% each and her homework grade counts for 20%. what is jessica's weighted average in the class rounded to the nearest whole number

1 answer:

Aleks04 [339]2 years ago

3 0

To solve the problem, get the percentage of each test by multiplying the score and the percentage then add it all up:

82 * .25 (highest test grade) + 65* .15 (lowest test grade) + 71*.20 (each test remaining) + 77*.20 (each test remaining) + 92*.20 (homework grade)

= 20.5 + 9.75 + 14.2 + 15.4 + 18.4 = 78.25 or 78% in whole number

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Answer:

The dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches

Step-by-step explanation:

We have that:

$Area = 128$

Let the dimension of the paper be x and y;

Such that:

$Length = x$

$Width = y$

So:

$Area = x * y$

Substitute 128 for Area

$128 = x * y$

Make x the subject

$x = \frac{128}{y}$

When 1 inch margin is at top and bottom

The length becomes:

$Length = x + 1 + 1$

$Length = x + 2$

When 2 inch margin is at both sides

The width becomes:

$Width = y + 2 + 2$

$Width = y + 4$

The New Area (A) is then calculated as:

$A = (x + 2) * (y + 4)$

Substitute $\frac{128}{y}$ for x

$A = (\frac{128}{y} + 2) * (y + 4)$

Open Brackets

$A = 128 + \frac{512}{y} + 2y + 8$

Collect Like Terms

$A = \frac{512}{y} + 2y + 8+128$

$A = \frac{512}{y} + 2y + 136$

$A= 512y^{-1} + 2y + 136$

To calculate the smallest possible value of y, we have to apply calculus.

Different A with respect to y

$A' = -512y^{-2} + 2$

Set

$A' = 0$

This gives:

$0 = -512y^{-2} + 2$

Collect Like Terms

$512y^{-2} = 2$

Multiply through by $y^2$

$y^2 * 512y^{-2} = 2 * y^2$

$512 = 2y^2$

Divide through by 2

$256=y^2$

Take square roots of both sides

$\sqrt{256=y^2$

$16=y$

$y = 16$

Recall that:

$x = \frac{128}{y}$

$x = \frac{128}{16}$

$x = 8$

Recall that the new dimensions are:

$Length = x + 2$

$Width = y + 4$

So:

$Length = 8 + 2$

$Length = 10$

$Width = 16 + 4$

$Width = 20$

To double-check;

Differentiate A'

$A' = -512y^{-2} + 2$

$A" = -2 * -512y^{-3}$

$A" = 1024y^{-3}$

$A" = \frac{1024}{y^3}$

The above value is:

$A" = \frac{1024}{y^3} > 0$

This means that the calculated values are at minimum.

<em>Hence, the dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches</em>

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Answer:

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Step-by-step explanation:

Given expression is,

$2\left(\log_3\left(8\right)+\log_3\left(z\right)\right)-\log_3\left(3^4-7^2\right)$

Now using logarithmic rule to solve the expression as follows,

Applying product rule of logarithmic,

$\log_c\left(a\right)+\log_c\left(b\right)=\log_c\left(ab\right)$

Therefore,

$2\log_3\left(8z\right)-\log_3\left(3^4-7^2\right)$

Applying power rule of logarithmic,

$a\log_c\left(b\right)=\log_c\left(b^a\right)$

Therefore,

$\log_3\left(\left(8z\right)^2\right)-\log_3\left(3^4-7^2\right)$

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2 years ago

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