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This is how you wrote your formula: 1200×(1 - (1 + 0.032/12 - 120))/(0.032/12), which is not correct. I believe it should be written like this: 1200 x (1 - (1+ 0.032/12)^-120) / (0.032/12) = $123,093.73 - This is the Present Value of 120 annuity payments of $1,200 each over a period of 10 years, or 120 months @ 3.2% comp. monthly.

thank you!

Thanks Rom. Here are the "combinations" listed:

{s, t, a, t} | {s, t, a, i} | {s, t, a, s} | {s, t, a, c} | {s, t, t, i} | {s, t, t, s} | {s, t, t, t} | {s, t, t, c} | {s, t, i, s} | {s, t, i, t} | {s, t, i, i} | {s, t, i, c} | {s, t, s, t} | {s, t, s, i} | {s, t, s, c} | {s, t, s, s} | {s, t, c, s} | {s, a, t, i} | {s, a, t, s} | {s, a, t, t} | {s, a, t, c} | {s, a, i, s} | {s, a, i, t} | {s, a, i, i} | {s, a, i, c} | {s, a, s, t} | {s, a, s, i} | {s, a, s, c} | {s, a, s, s} | {s, a, c, s} | {s, i, s, t} | {s, i, s, i} | {s, i, s, c} | {s, i, s, s} | {s, i, t, i} | {s, i, t, c} | {s, i, t, s} | {s, i, i, c} | {s, i, i, s} | {s, i, c, s} | {s, s, t, i} | {s, s, t, c} | {s, s, t, s} | {s, s, i, c} | {s, s, i, s} | {s, s, c, s} | {t, a, t, i} | {t, a, t, s} | {t, a, t, t} | {t, a, t, c} | {t, a, i, s} | {t, a, i, t} | {t, a, i, i} | {t, a, i, c} | {t, a, s, t} | {t, a, s, i} | {t, a, s, c} | {t, a, s, s} | {t, a, c, s} | {t, t, i, s} | {t, t, i, t} | {t, t, i, i} | {t, t, i, c} | {t, t, s, t} | {t, t, s, i} | {t, t, s, c} | {t, t, s, s} | {t, t, t, i} | {t, t, t, c} | {t, t, t, s} | {t, t, c, s} | {t, i, s, t} | {t, i, s, i} | {t, i, s, c} | {t, i, s, s} | {t, i, t, i} | {t, i, t, c} | {t, i, t, s} | {t, i, i, c} | {t, i, i, s} | {t, i, c, s} | {t, s, t, i} | {t, s, t, c} | {t, s, t, s} | {t, s, i, c} | {t, s, i, s} | {t, s, c, s} | {a, t, i, s} | {a, t, i, t} | {a, t, i, i} | {a, t, i, c} | {a, t, s, t} | {a, t, s, i} | {a, t, s, c} | {a, t, s, s} | {a, t, t, i} | {a, t, t, c} | {a, t, t, s} | {a, t, c, s} | {a, i, s, t} | {a, i, s, i} | {a, i, s, c} | {a, i, s, s} | {a, i, t, i} | {a, i, t, c} | {a, i, t, s} | {a, i, i, c} | {a, i, i, s} | {a, i, c, s} | {a, s, t, i} | {a, s, t, c} | {a, s, t, s} | {a, s, i, c} | {a, s, i, s} | {a, s, c, s} | {i, s, t, i} | {i, s, t, c} | {i, s, t, s} | {i, s, i, c} | {i, s, i, s} | {i, s, c, s} | {i, t, i, c} | {i, t, i, s} | {i, t, c, s} | {i, i, c, s}=125 

This is what I found.

If no ordering matters at all then 34 is the correct answer.

I'm not sure what ordering they consider distinct to come up with 125

your mom

Thank you very much Rom. I greatly appreciate your time. However, the answers given to us are:

Permutations = 386,  Combinations = 125. Both of them are supposed to be "Distinct". But, as of yet, no detailed explanations as to how they were arrived at!. They are also listed in some sort of "lexicographic" order, which I have.

Wow thanks your explanations help so much:)

Is "nosceince" supposed to be "nonsense" ?

possibly 1, but probably not

definitely 2

definitely 3

not 4

\(\text{This one is the reverse of the others}\\ \text{Let }w \text{ be the weight you're after in this problem}\\ 0.2 = \Phi\left(\dfrac{261-w}{9.4}\right)\)

\(\Phi^{-1}(0.2) = \dfrac{261-w}{9.4}\\ \text{So use your table or software or whatever to look up the argument, }x,\\ \text{to }\Phi(x) \text{ that corresponds to }\Phi(x) = 0.2\\ \text{Once you've got that value simple algebra will solve for }w\)

\(\text{similar to last one}\\ p = \Phi\left(\dfrac{37-33}{2}\right)-\Phi\left(\dfrac{29-33}{2}\right)\\ n = 400 p\)

\(\text{Let }\Phi(x) \text{ be the CDF of the standard normal distribution}\\ P[\text{weight > 165}] = 1 - P[\text{weight < 165}] = \\ 1 - \Phi\left(\dfrac{165-176}{11.4}\right)\approx \\ 1 - \Phi(-0.964912) =p \text{ (meaning you figure out what }p \text{ is)}\)

\(p\cdot 120 = ?\\ \text{match what you get above with the choice list}\)

nice

sorry, you wer rite but i thot it was to hard. sorry.