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marusya05 [52]
3 years ago
8

As the administration looks ahead to next year, Kyra is concentrating on finishing up this year well. Students at her school rec

eive 4 quarter grades, a mid-term exam grade, and a final exam grade for each class. The quarter grades are all weighted equally, the mid-term exam grade is double the weight of a quarter grade, and the final exam grade is four times the weight of a quarter grade. What fraction of the final course grade is the final exam grade? Express your answer as a common fraction.
Mathematics
1 answer:
kakasveta [241]3 years ago
8 0

Answer:

2/5

Step-by-step explanation:

Defining <em>x </em>the weight of each quarter grades (4x for the 4 quarter grades), then the weight of the mid-term exam grade is 2x, and the weight of the final exam is 4x. The addition of all grades is 4x + 2x + 4x = 10x (this represents the combination of grades that makes the final course grade).

So, the final exam grade represents 4x/10x = 2/5 of the final course grade.

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Solve the given linear Diophantine equation. Show all necessary work. A) 4x + 5y=17 B)6x+9y=12 C) 4x+10y=9
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Answer:

A) (-17+5k,17-4k)

B)  (-4+3k,4-2k)

C) No integer pairs.

Step-by-step explanation:

To do this, I'm going to use Euclidean's Algorithm.

4x+5y=17

5=4(1)+1

4=1(4)

So going backwards through those equations:

5-4(1)=1

-4(1)+5(1)=1

Multiply both sides by 17:

4(-17)+5(17)=17

So one integer pair satisfying 4x+5y=17 is (-17,17).

What is the slope for this equation?

Let's put it in slope-intercept form:

4x+5y=17

Subtract 4x on both sides:

     5y=-4x+17

Divide both sides by 5:

      y=(-4/5)x+(17/5)

The slope is down 4 and right 5.

So let's show more solutions other than (-17,17) by using the slope.

All integer pairs satisfying this equation is (-17+5k,17-4k).

Let's check:

4(-17+5k)+5(17-4k)

-68+20k+85-20k

-68+85

17

That was exactly what we wanted since we were looking for integer pairs that satisfy 4x+5y=17.

Onward to the next problem.

6x+9y=12

9=6(1)+3

6=3(2)

Now backwards through the equations:

9-6(1)=3

9(1)-6(1)=3

Multiply both sides by 4:

9(4)-6(4)=12

-6(4)+9(4)=12

6(-4)+9(4)=12

So one integer pair satisfying 6x+9y=12 is (-4,4).

Let's find the slope of 6x+9y=12.

6x+9y=12

Subtract 6x on both sides:

      9y=-6x+12

Divide both sides by 9:

       y=(-6/9)x+(12/9)

Reduce:

       y=(-2/3)x+(4/3)

The slope is down 2 right 3.

So all the integer pairs are (-4+3k,4-2k).

Let's check:

6(-4+3k)+9(4-2k)

-24+18k+36-18k

-24+36

12

That checks out since we wanted integer pairs that made 6x+9y=12.

Onward to the last problem.

4x+10y=9

10=4(2)+2

4=2(2)

So the gcd(4,10)=2 which means this one doesn't have any solutions because there is no integer k such that 2k=9.

6 0
3 years ago
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