#how do i do these problems

10 messages · Page 1 of 1 (latest)

hallow lichen
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i just did an online test thingy, and i screenshotted every problem i couldnt/didnt do (dw it ended and i submitted everything i had) ill paste them in below

vestal pelicanBOT
hallow lichen
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those are the problems i couldnt do

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i might have been able to do some of them but i was running out of time at the end

knotty magnet
# hallow lichen

I cant really help with the first and second problem tbh but ill try to help with the other 3

For the third one how id do it is id set up an equation
First split it up into 2 cases:
The first case is that you pull a yellow marble from the first bag, the probability of this happening is 80% and for you to pull the same marble out of both bags youd need to pull a yellow marble from the second bag too. this has a 16/(16+x) chance of happening where x is the number of red marbles in the second bag
The second case is that you pull a red marble from the first bag. the probability of this hpapening is 20% and for you to pull the same marble out of both bags youd need to pull a red marble from the second bag too. this has an x/(16+x) chance of happening where x is once again the number of red marbles in the second bag

So the probability of you pulling the same marble from both bags through the first case is
80% * 16/(16+x) (the probability of the case happening multiplied by the probability of yoou pulling the same marble from both bags when it happens)
The probability of you pulling the same marble from both bags through the second case is
20% * (x/(16+x)) (the probability of the case happening multiplied by the probability of you pulling the same marble from both bags when it happens)

So the total probability of you pulling the same colored marble from both bags is the probability of it happening through case A plus the probabliity of it happening through case B
which is
80%(16/(16+x)) + 20%(x/(16+x))
And as the question has told us the probability of pulling the same colored marble from both bags is 1/4, so we'll set this equal to 1/0%(16/(16+x)) + 20%(x/(16+x)) = 1/4
And then just solve for x to get the number of red marbles in the second bag

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For the fourth one:
You need to use the fact that the partial sum of the first n elements minus the sum of the first (n-1) elements is just the nth element (try to use intuition to see why this is true, maybe on a simpler sequence like the sequence 1,2,3,4,5...)
So we'll use this fact to get that the nth term = (n^2 -3n) - ((n-1)^2 - 3(n-1))
Then expand and simplify to get a simpler expression for the nth term and plug in n = 2025

knotty magnet
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For the 5th one i kinda got stuck after a certain point tbh and since i dont wanna spend more than 30 minutes on it tbh ill just send what ive gotten till now
youll need to use some properties of logarithms
First take the exponent out of the logarithm to get that
x * log base x of y = 9
and
5y * log base y of x = 225
divide the second equation by 5 to simplify it to
y * log base y of x = 45
Then youll need to use the property that log base x of y = (logy)/(logx)
youll apply this to both equations to get that
x * (logy)/(logx) = 9
and
y * (logx)/(logy) = 45
then youll slightly alter the form of the first equation to make it become
x * 1/( (logx)/(logy) ) = 9
then solve for (logx)/(logy) in terms of purely y by dividing both sides of the second equation by y
so it goes from
y * (logx/logy) = 45
to
(logx)/(logy) = 45/y
so substitute that into the first equation to get
x * 1/( 45/y ) = 9
simplify to get that
xy/45 = 9
so
xy = 5
y = 5/x
From here i dont really know where to go tbh but maybe youll be able to proceed from here
I feel like it might end up being really obvious but writing out all of the solutions kinda exhausted my brain so maybe thats why im not being able to find it

hallow lichen
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