Algebra question?

Question

An arithmetic student needs an average of 70 or more to receive credit for the course. She scored 82, 64, and 98 on the first three exams. Write an inequality representing the score she must get on the last test to receive credit for the course.

Answer 1

Assuming the exams have equal weight it would be a straight average of the four tests.

Let n be the required score on the last test:
(82 + 64 + 98 + n) / 4 ≥ 70

Now you need to solve for n. Start by multiplying through by 4 on both sides:
82 + 64 + 98 + n ≥ 280

Now add up the numbers on the left:
244 + n ≥ 280

Now subtract 244 from both sides:
n ≥ 36

The student must score a 36 or better on the final exam... seems pretty doable given her prior scores.

Answer 2

Remember that the average of 4 tests is calculated with the following:

(T1 + T2 + T3 + T4)/4 = A

Where A is the average and T1 through T4 are the scores.

Since she needs at least 70, and we know what T1 through T3 equal, then

(82 + 64 + 98 + T4)/4 >= 70

Translation: The average of her test scores must be greater than or equal to 70. Now we solve this as if it were an equation.

(244 + T4)/4 >= 70
(244 + T4) >= 280
T4 >= 36

Therefore, she must get at least 36 to receive credit for the course on her fourth test.

Let's check:

(T1 + T2 + T3 + T4)/4 = (82 + 64 + 98 + 36)/4 = 70

Answer 3

82+64+98+X = 70x4
244+X = 280
X=36

Answer 4

X = The minimum score we need in the last exam

Step 1: 82/4+64/4+98/4+X/4 = 70

Step 2: 244+ X = 280

Step 3 : X = 36 !

:-)

Answer 5

82+64+98=244 so they will need to get a score over fourty and they will be fine they will get an average of a 70 or high if the score above 40

Answer 6

(82+64+98+x)/4>=70

Answer 7

(82 + 64 + 98+ x)/4 >= 70
x >= 36

Answer 8

sum must be greater than 280.

Therefore,
x+82+64+98>=280

Therefore x>=36

Answer 9

suppose that the mark she has to get = x
so
(82 + 64 + 98 +x)/4 ≥ 70................*4
224 + x ≥ 280
x ≥ 280 -224
x ≥ 36
so she must get 36 or more

Answer 10

For increasing cubic binomials the final formula is as follows: (a + b) ^ 3 = a^3 + 3*a^2*b^a million + 3*a^a million*b^2 + b^3 on your case, a is x and b is -y^5 So (x - y^5)^3 = x^3 + 3*x^2*(-y^5)^a million + 3*x^a million*(-y^5)^2 + (-y^5)^3 Simplified: =x^3 - 3x^2*y^5 + 3x*y^10 - y^15 :D