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2 years ago

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In a given course a student receives preliminary examination grades of 81, 85, and 95. The final examination is weighed for one-

third and the average of the preliminary grades is weighed as 2/3 of the final grade. What should the final examination grade be for a semester average of 90?
99
96
93
88
87

1 answer:

Vsevolod [243]2 years ago

4 0

Answer:

B. 96

Step-by-step explanation:

Let x represent score of final third exam.

We have been given that in a given course a student receives preliminary examination grades of 81, 85, and 95.

Let us find average of preliminary scores.

$\frac{81+85+95}{3}=\frac{261}{3}=87$

We have been given that the final examination is weighed for one-third and the average of the preliminary grades is weighed as 2/3 of the final grade. This means that final grade is equally weighed into 3 parts. The two parts are 87 each and third part is x.

Using our given information, we can set an equation as:

$\frac{87+87+x}{3}=90$

$\frac{174+x}{3}=90$

$\frac{174+x}{3}*3=90*3$

$174+x=270$

$174-174+x=270-174$

$x=96$

Therefore, the final examination grade should be 96 for a semester average of 90.

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Given directed line segment QS , find the coordinates of R

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Answer:

The answer is below

Step-by-step explanation:

The question is not complete, what are the coordinates of point Q and R. But I would show how to solve this.

The location of a point O(x, y) which divides line segment AB in the ratio a:b with point A at ( $x_1,y_1$ ) and B( $x_2,y_2$ ) is given by the formula:

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If point Q is at ( $x_1,y_1$ ) and S at ( $x_2,y_2$ ) and R(x, y) divides QS in the ratio QR to RS is 3:5, The coordinates of R is:

$x=\frac{3}{3+5}(x_2-x_1)+x_1=\frac{3}{8}(x_2-x_1)+x_1\\ \\y=\frac{3}{3+5}(y_2-y_1)+y_1=\frac{3}{8}(y_2-y_1)+y_1$

Let us assume Q(−9,4) and S(7,−4)

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Find the mass and center of mass of the lamina that occupies the region D and has the given density function rho. D = {(x, y) |

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Answer:

$M=168k$

$(\bar{x},\bar{y})=(5,\frac{85}{28})$

Step-by-step explanation:

Let's begin with the mass definition in terms of density.

$M=\int\int \rho dA$

Now, we know the limits of the integrals of x and y, and also know that ρ = ky², so we will have:

$M=\int^{9}_{1}\int^{4}_{1}ky^{2} dydx$

Let's solve this integral:

$M=k\int^{9}_{1}\frac{y^{3}}{3}|^{4}_{1}dx$

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So the mass will be:

$M=21k*8=168k$

Now we need to find the x-coordinate of the center of mass.

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Now we need to find the y-coordinate of the center of mass.

$\bar{y}=\frac{1}{M}\int\int y*\rho dydx$

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$\bar{y}=\frac{255}{672}8=\frac{2040}{672}$

$\bar{y}=\frac{85}{28}$

Therefore the center of mass is:

$(\bar{x},\bar{y})=(5,\frac{85}{28})$

I hope it helps you!

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