Simplification of 4[(.5)^4 * (.6)^3]+4[(.5)^4 * (.6)^3 * (.4)] = 4(.3)^3 * (.7)
- Thread starter bllove07
- Start date
- Joined
- Apr 12, 2005
- Messages
- 11,339
It may help to write everything in fractional form.
.5 = 1/2
.6 = 3/5
.4 = 2/5
.3 = 3/10
.7 = 7/10
Actuaries are trained to wade through the nitty-gritty details. That is all that is needed, here.
Steven G
Elite Member
- Joined
- Dec 30, 2014
- Messages
- 14,561
4[(.5)^4 * (.6)^3]+4[(.5)^4 * (.6)^3 * (.4)]
To the left of the + we are multiply 4, .5^4 and .6^3. To the right of the + we are multiplying 4, .5^4, .6^3 and .4
What is underlined is common so it can be factored out yielding
(4 * .5^4 * .6^3)(1 + .4)= (4 * .5^4 * .6^3)(1.4) = (4 * .5*.5^3 * .6^3)(1.4) = (4 * .5 * .5^3 * .6^3)(1.4) = (4 * .5 *.3^3)(1.4) = (4 * .5 *.3^3 * 1.4) = (4 *.3^3 * .7).
Have fun passing your exams. I passed two of them. They were not too bad if you are prepared.
Steven G
Elite Member
- Joined
- Dec 30, 2014
- Messages
- 14,561
In my opinion there is no left side. OP needs to convert the lhs to the rhs. Fair enough, he can use the rhs but for beginners he should just work on with what is given (the lhs) and arrive with the rhs.Get rid of the 3 "4"s (divide each side by 4):
.5^4 * .6^3 + .5^4 * .6^3 * .4 = .3^3 * .7
Let u = .5^4 * .6^3:
u + .4u = .3^3 * .7
Play with that
BTW, do you have a patent on the u, v substitution?
I shall contest it if he does.