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<average and average <100
or
90<average<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<(76+99+86+x)/4
times 4 both sides
360<261+x
minus 261 both sides
99<x

(76+99+86+x)/4<100
times 4 both sides
261+x<400
minus 261 both sides
x<139

so

99<x<139
since max score is 100
99<x<100

interval notaion is
[99,100]

answer is 2nd one

3 0

2 years ago

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The 18 faculty members in a college math department range in age from 32 to 68. A stem plot follows: The first and third quartil

valentina_108 [34]

Answer:

$Q_{1}=39, \:and\: Q_{3}=58$

Step-by-step explanation:

1) This Stem and Leaf plot works like a Histogram. Completing the question, check below the Stem plot. Look the data out of the Stem plot and within (in the graph below)

32, 34 , 38 , 39 , 39 , 40 , 43 , 45 , 46 , 49 , 53 , 54 , 57 , 58 , 59 , 59 , 63 , 68

2) Notice how the first digit is on the Stem column (check below)

3) To find the first and Third Quartiles

Let's use the formulas to find the position, then the value. As it follows:

$Q_3=\frac{3}{4}(n+1) \Rightarrow Q_3=\frac{3}{4}(18+1) Q_3=14.25 \approx14th =58$