Please help with two- Way Frequency tables (picture attached)

Let's start with the column and row totals, which should make things easier

Total <$20K = 69+112+102+13 = 296

total $20-30K  500

total >30K   404

TOTAL 296+500+404=1200

Rows,

<HS   69+36+2 = 107

HS 224

Some C  438

C     431

Check total: 107+224+438+431 = 1200, good

A. P(<HS)  = 107/1200

because there are 107 folks who didn't finish high school out of the 1200 total.

B.  P(<$20K) = 296/1200 = 37/150

C.  P(>$30K & <HS)

Only 2 of 1200 folks fit this description, upper right box.

P(>$30K & <HS) = 2/1200 = 1/600

D.  P(<$20K & C) = 13/1200

That's the bottom left box.

E. P(>$30K & HS) = 14/1200 = 7/600

F. P(<HS OR >$30K)

OR is usually more complicated.  We have 107 who didn't finish high school and 404 more than $30K.   We counted the 2 in the box <HS AND >$30K twice, so our total is 107 + 404 - 2 = 509

P(<HS OR >$30K) = 509/1200

answer: standard form is ax^2 + bx + c = 0

step-by-step explanation:

standard form is ax^2 + bx + c = 0

(2x + 1)(x - 2) = 0

distribute to get into standard form

2x^2 - 4x + x - 2 = 0

2x^2 - 3x - 2 = 0

a = 2, b= -3, c = -2

a is your answer

step-by-step explanation: