so we have the points of (0,-7),(7,-14),(-3,-19), let's plug those in the y = ax² + bx + c form, since we have three points, we'll plug each one once, thus a system of three variables, and then we'll solve it by substitution.
well, from the 1st equation, we know what "c" is already, so let's just plug that in the 2nd equation and solve for "b".
well, now let's plug that "b" into our 3rd equation and solve for "a".
![\bf -19=9a-3b-7\implies -12=9a-3b\implies -12=9a-3(-1-7a) \\\\\\ -12=9a+3+21a\implies -15=9a+21a\implies -15=30a \\\\\\ -\cfrac{15}{30}=a\implies \blacktriangleright -\cfrac{1}{2}=a \blacktriangleleft \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{and since we know that}}{-1-7a=b}\implies -1-7\left( -\cfrac{1}{2} \right)=b\implies -1+\cfrac{7}{2}=b\implies \blacktriangleright \cfrac{5}{2}=b \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill y=-\cfrac{1}{2}x^2+\cfrac{5}{2}x-7~\hfill](https://tex.z-dn.net/?f=%5Cbf%20-19%3D9a-3b-7%5Cimplies%20-12%3D9a-3b%5Cimplies%20-12%3D9a-3%28-1-7a%29%20%5C%5C%5C%5C%5C%5C%20-12%3D9a%2B3%2B21a%5Cimplies%20-15%3D9a%2B21a%5Cimplies%20-15%3D30a%20%5C%5C%5C%5C%5C%5C%20-%5Ccfrac%7B15%7D%7B30%7D%3Da%5Cimplies%20%5Cblacktriangleright%20-%5Ccfrac%7B1%7D%7B2%7D%3Da%20%5Cblacktriangleleft%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Band%20since%20we%20know%20that%7D%7D%7B-1-7a%3Db%7D%5Cimplies%20-1-7%5Cleft%28%20-%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%29%3Db%5Cimplies%20-1%2B%5Ccfrac%7B7%7D%7B2%7D%3Db%5Cimplies%20%5Cblacktriangleright%20%5Ccfrac%7B5%7D%7B2%7D%3Db%20%5Cblacktriangleleft%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20y%3D-%5Ccfrac%7B1%7D%7B2%7Dx%5E2%2B%5Ccfrac%7B5%7D%7B2%7Dx-7~%5Chfill)
I believe that is C. Statistical question.
I know it’s not B because that has nothing so do with the question so you’re good there. While A could be a potential answer, it says a question that can be answered by collecting data, whereas the mean would actually be that data and not the question that can be answered so C would be the only choice left.
9 is 1/10 of 90 because if you add 1 zero it is 90
Answer: 0.5143
Step-by-step explanation:
Probability of students who are over 21 years old = 30% = 0.3
Probability of students who are under 21 years old = 100% - 30% = 70% = 0.7
Probability of drinking alcohol for over 21 = 80% = 0.8
Probability of not drinking alcohol for over 21 = 100%- 80% = 20% = 0.2
Let the probability of the students who are not over 21, that drink alcohol be p.
Total probability of a college student drinking alcohol = (0.3 × 0.8) + (0.7 × p)
0.6 = (0.3 × 0.8) + (0.7 × p)
0.6 = 0.24 + 0.7p
0.7p = 0.6 - 0.24
0.7p = 0.36
p = 0.36/0.7
p = 0.5143
The probability of the students who are not over 21, that drink alcohol is 0.5143.
I rather not..............
