Going into the final exam, which will count as two test, brooke has a test score of 75,84,74,59, and 94. what score does brooke need on the final in order to have and average score of an 80?

There have been 5 tests so far.

The final will count for two more.

So the grand average score will be the average of 7 test scores.

If the grand average score is to be 80, then all 7 components

will add up to (7 x 80) = 560 .

The tests she has already taken add up to

(75 + 84 + 74 + 59 + 94) = 386

So the "two" tests she's about to take need to add at least

(560 - 386) = 174 .

So she needs to score at least (174/2) = 87 on the final.

It'll be averaged into the others as if it were two tests,

so it'll add 174 points to the others, and the whole set

of ' 7 ' tests will average 80 .

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3