What are th minimum, first quartile, median, third quartile, and maximum of the data set? 63,98,40,32,20,80,102,65

First quartile = 36 , Median = 64 , Minimum  = 20 ,

Maximum = 102, Third quartile = 89.

Step-by-step explanation:

Given  : 63,98,40,32,20,80,102,65.

To find : What are the minimum,first quartile,median,third quartile,and maximum of the data set.

Solution : We have given that 63,98,40,32,20,80,102,65.

Arranging in ascending order

20 , 32 , 40 ,63 ,65 ,80 ,98 , 102.

Median =   = 64.

Minimum  = 20 .

Maximum = 102.

First quartile =   [tex]\frac{32 +40}{2} = 36.

Third quartile =   [tex]\frac{80 +98}{2} = 89.

Therefore, First quartile = 36 , Median = 64 , Minimum  = 20 ,

Maximum = 102, Third quartile = 89.

37/3

step-by-step explanation:

step 1: simplify both sides of the equation.

7x+8(x+

1

4

)=3(6x−9)−8

7x+(8)(x)+(8)(

1

4

)=(3)(6x)+(3)(−9)+−8(distribute)

7x+8x+2=18x+−27+−8

(7x+8x)+(2)=(18x)+(−27+−8)(combine like terms)

15x+2=18x+−35

15x+2=18x−35

tep 2: subtract 18x from both sides.

15x+2−18x=18x−35−18x

−3x+2=−35

step 3: subtract 2 from both sides.

−3x+2−2=−35−2

−3x=−37

step 4: divide both sides by -3.

−3x

−3

=

−37

−3

x=

37

3

to determine the unit cost your would divide the total cost by the number sold.

7 bagels for $4.41       $4.41/7 = $.63 each at fresh side

5 bagels for $3.60         $3.60/5 = $.72 each at value mart