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:
59
Step-by-step explanation:
We know that all the angles must add up to 180
so we have
38+2b-20+b-15=180
solve for b
3b+3=180
3b=177
b=59
Recheck the answer with your teacher.
OK. I did it, and I have the solution.
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The length of the deck is (5 + 2x) .
The width of the deck is (4 + 2x) .
If the deck didn't have that big hole in the middle where the pool is,
then its area would be
(5 + 2x) · (4 + 2x) .
When you multiply that all out, you get Area = 4x² + 18x + 20
and the question tells us that the area of the whole big rectangle is 90 yds² .
So we can write
4x² + 18x + 20 = 90 .
Subtract 90 from each side: 4x² + 18x - 70 = 0
Divide each side by 2 : 2x² + 9x - 35 = 0
You can use the quadratic equation to solve that and find out that
x = 2.5 yards, and that's what the question is asking you.
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That makes the deck 10 yds high and 9 yds wide.
Total area of the whole big rectangle, (deck + pool ), = 90 yds².
