This is a classic probability question. Once you know that one coin is a head, there are 3 possible outcomes for the two coins with at least one head.
HH
HT
Q: How many 6 digit numbers are there whose digits sum to 51?
A: 56.
There are 3 possible sets of 6 numbers: 999996 where the 6 can be the 1st, 2nd, 3rd, 4th, 5th, 6th place values making 6 6 digit numbers
3 9s and 3 8s which is the same probability as 3 boys and 3 girls with 6 children 3C6 = 20 6 digit numbers
and 4 9s with an 8 and a 7 with 30 possibilities
789999
798999
799899
799989
799998
879999
897999
899799
899979
899997
978999
979899
979989
979998
987999
989799
989979
989997
997899
997989
997998
998799
998979
998997
999789
999798
999879
999897
999978
999987
credit to Julia Robinson Mathematics Festival

This question asks for absolute value which can be found either in the Math menu or using catalog (2nd, zero) and abs comes up right away because it begins with an A :)




Domain means x values and Range means y.
D comes before R in the alphabet and x is before y.
This problem gives us x values so use them to find y.
f(x) = 2x^2 – 8
f(-2) = 2(-2)^2 – 8 = 2(4) – 8 = 0 (This narrows it down to answers (3) or (4)!)
f(3) = 2(3)^2 – 8 = 2(9) – 8 = 10 (This makes the answer jump out at ya!)
f(5) = 2(5)^2 – 8 = 2(25) – 8 = 42
Can also use the table from the TI-84 :)


How did 6 – 2(x^2 – 1) become 6 – 2x^2 +2?
Help Evan identify this first step of his problem solving :)
That’s it! 2 points on the Regents!
