AVERAGE between 90 and 100%
means not including 90 and 100% but not including 90 or 100%
average=(sum of values in set)/(how many values in the set)
so
4 exams, so there are 4 values
average=(sum)/4
the sum is the known+unknwon
sum=76+99+86+x
so
90<span><</span>average and average <span><</span>100
or
90<span><</span>average<span><</span>100
lets say average=a
a=sum/number
a=(76+99+86+x)/4
90<(76+99+86+x)/4<100
90<(76+99+86+x)/4 and (76+99+86+x)/4<100
solve each for x and find intersection
90<span><</span>(76+99+86+x)/4
times 4 both sides
360<span><</span>261+x
minus 261 both sides
99<span><</span>x
(76+99+86+x)/4<u><</u>100
times 4 both sides
261+x<u><</u>400
minus 261 both sides
x<u><</u>139
so
99<u><</u>x<u><</u>139
since max score is 100
99<u><</u>x<u><</u>100
interval notaion is
[99,100]
answer is 2nd one
Answer:
-3v
Step-by-step explanation:
-6v is a negative when 3v is positive
when you add 3v to -6v it equals -3v since 6 is greater
Answer:
No solution
Step-by-step explanation:
4x+9=4x+13
9 ≠ 13
No solution
In this problem you will need to use the Pythagorean theorem (c^2=a^2+b^2).
The a and b represents the two edges, while c is the diagonal side and it is called the hypotenuse. Since you already know what the hypotenuse is and what one of the sides already are you just have to use the problem: c^2-a^2=b^2. Then if you plug the data you already have into the problem you will get 10^2-6^2=b^2. That then equals 100-36=b^2. Then you subtract and get b^2=64. Then you square root both sides and you get the answer b=8.
8,500,000
Round the number 4 to 5 because the number 7 is behind it.
