should i use the residual plot in comparing the industrailized countries vs. less-developed country? i mean, is it relevant if i use that?
coz here is my thesis...What are the effects of population, GDP, GDP per capita and unemployment rate in the economy of industrialized country and...
guys, is it like this...?
for t2 = nC1 (ax)^n-1 (by)^1 = 34992x^n-1y
bna^n-1 = 34992
a^n = 6561
a^n = 3^8
therefore the answer for n = 8 and for a = 3, so you just have to find the b. but the second equation is bna^n-1 = 34992, so ill just substitute the values that i found...
it is like...
guys, is it like this...?
for t2 = nC1 (ax)^n-1 (by)^1 = 34992x^n-1y
bna^n-1 = 34992
a^n = 6561
a^n = 3^8
therefore the answer for n = 8 and for a = 3, so you just have to find the b. but the second equation is bna^n-1 = 34992, so ill just substitute the values that i found...
it is like...
anyone could help me with this question...
in the expansion of (ax + by)^n, the coeffiients of the first 3 are 6561, 34992, and 81648., Find the value if a, b, and n.
i did this...
t1 = nC0 (ax)^n = 6561x^n
a^n = 6561
t2 = nC1 (ax)^n-1 (by)^1 = 34992x^n-1y
bna^n = 34992
but i'm not...
this is the problem...
there are 25 students in a Data Management Class. Determine the probability that at keast two of them share the same birthday. (Assume that every year has 365 days)
--> i have no clue what technique what to do here... coz this is a thinking question so its kinda hard for...
this is the problem...
1. nine horses are entered in a horse race. If you "box" three horses (three are chosen and they can finish in any of the first 3 positions in the race), determine the probability that you will hold the winning ticket.
-->i tried this one... (3C1*3C2*3C3)/9C3 but my...
tide, i still cannot get the right answer. like i solve many times, but i really can't find the correct answer... i tried this one... 5C4*51*1/52C4, but its is wrong since the right answer in my book is 1/20825
i have no idea which technique of probability to choose. here is the problem...
1. a five-card poker hand is dealt and the last card is turned face up. Determine the probability that you have been dealt four aces, given that the card turned over is an ace.
2. I how many ways can a group of...
i got like (2n+4)(2n+3)(2n+2) = 2/3(n+4)(n+3)(n+2)(n+1)
then... 2(n+2)(2n+3)2(n+1) = 2/3(n+4)(n+3)(n+2)(n+1)
which is 4(2n+3) = 2/3(n+4)(n+3)
and i dont know wat to do next
what if you get like this.... -24(2n+3)(n+4)(n+3) = 0...
what should i do to the -24? or it is just 0, then the value for n is just n=-2/3, n=-4 and n=-3?
correct me if im wrong ok? i've try the other side... should you make the numerator 2/3(n+4)! to -2/3(n-4)! in order to cancel it from n(n-1)(n-2)(n-3)(n-4)?
ceptimus, is my work on the right track?
is it 9!/5! or 9x8x7x6 for 4d? since the numbers runs from 0 to 9, but there is only 5 odd numbers, and same thing with 4e?
this is the question...
4. How many four digit numbers are there with the following retrictions?
d) the number is odd
e) the...