Answer:
You dind't include the answer choices but it should look something like

Mid term :
Q1 = (88 + 85)/2 = 86.5
Q2 = (92 + 95)/2 = 93.5
Q3 = 100
IQR = Q3 - Q1 = 100 - 86.5 = 13.5
final exams :
Q1 = (65 + 78)/2 = 71.5
Q2 = (88 + 82)/2 = 85
Q3 = (95 + 93)/2 = 94
IQR = Q3 - Q1 = 94 - 71.5 = 22.5
so the final exams has the largest IQR
Answer:
EF = 58/7
Step-by-step explanation:
If you extend both BC and AD to meet at P you get 3 similar triangles PAB, PEF, and PCD. Let's call BF = 3x and FC = 4x. Therefore the following proportions can be expressed:
BP/7 = (BP+7x)/10
10BP = 7*(BP+7x)
10BP = 7BP+49x
3BP = 49x
BP = (49/3)x
BP/7 = (BP+3x)/EF
EF = (BP+3x)*7/BP
EF = ((49/3)x + 3x)*7/((49/3)x)
EF = ((58/3)x*7*3)/(49x)
EF = 58/7
Y *x =20
y +x =9
rewrite 1st eq. as y = 20/x
subsituute that into 2nd eq.
20/x +x =9
multiply each term by X
20 +x^2 = 9x
subtract 9x from each side:
x^2 - 9x +20 = 0
now factor:
(x-5) (x-4) = 0
solve for x
5-5 = 0
4-4 = 0
so X can be 4 or 5
replace x into equations and solve for Y
so when X = 4, Y = 5
When X = 5, Y = 4
x= 5,4
y = 4,5
Answer:
y=-2x-6
Step-by-step explanation: