29293
Step-by-step explanation:
yes yes !!
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
<span>the are 8 sixes in the number 48
Hope I helped
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Your question: Find D in | - 2d|<10
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SOLVING....
Solve the absolute value
|−2d| < 10
=
−2d < 10
or
−2d > −10
Solve −2d < 10 for D
Divide each side by -2
-2d ÷ -2 < 10 ÷ -2
d > -5
Now Solve −2d > −10
Divide each side by -2
-2d ÷ -2 > -10 ÷ -2
d < 5
So D is d < 5 and d > -5