Bonita had a collection of 804 beads. She gave 283 to a friend which is the closest estimate of the number of beads she has left

delaney would like to make a 5lb nut mixture that is 60% peanuts and 40% almonds.

she has several pounds of peanuts and several pounds of a mixture that is 20% peanuts and 80% almonds.

let 'p' represent the number of pounds of peanuts needed to make the new mixture, and  

let 'm' represent the number of pounds of the 80% almond-20% peanut mixture.

:

p + m = 5

therefore

m = (5-p)

write an equation based on the percent of peanuts

1p + .2m = .6(5)

replace m with (5-p)

p + .2(5-p) = .6(5)

p + 1 - .2p = 3

p - .2p = 3 - 1

.8p = 2

p = 2/.8

p = 2.5 lbs of peanuts required

then

5 - 2.5 = 2.5 lbs of the peanut/almond mixture required

:

a) what is the system that models this situation?

p + m = 5

1p + .2m = .6(5)

the answer is: -22 also a

hope i

-lavira: )

school a:

minimum value is 2

maximum value is 22

the lower quartile is 2.5

the median is 10

the upper quartile is 15.5

school b:

minimum value is 9

maximum value is 20

the lower quartile is 12

the median is 16

the upper quartile is 18

box plot for school a isn't symmetrical. the data tails on the right

box plot for school b isn't symmetrical. the data tails on the left

hope i

-lavira: )