Mathematics Question Thread

[RuiAce]

ok thanks! i was kinda implying that, but i didnt state it explicitly should i do that next time

also answer to question is attached

Yeah it's highly recommended because assuming what you're trying to prove is technically classified as a mistake. Whilst in high school maths I would probably just ignore it, some harsh markers can pick you out for it. Best be safe than sorry

[amelia20181]

how do you do

[Ali_Abbas]

how do you do

Recall that \(\sqrt{a}\) = \(a^{1/2}\) for all non-negative real numbers \(a\). Simply make this substitution into the integrand and multiply the powers, giving \(5/2\) as the new exponent. You now have something of the form \(\int(ax+b)^{n}dx\) which will be equal to \(\frac{1}{a(n+1)}(ax+b)^{n+1} + C\).

[amelia20181]

how do you do

[Fergus6748]

how do you do

Hey, so the way I would do the question would be to first convert it to index form, so: 2(x-3)^-2, and from there integrate to get -2(x-3)^-1 or -2/(x-3). That would be when you use your limits, x=0,1.
f(x)=(-2/(1-3))-(-2/(0-3))
     =(1-2/3)
so the answer would be:
     =1/3 units^2
im pretty sure

[LaraC]

Hello!
Could someone please help me with the following question:
In a set of 30 cards, each one has a number on it from 1 to 30. If 1 card is drawn out, then replaced and another drawn out, find the probability of getting:
a) a 3 on the first card and an 18 on the second card (I have this answer --> 1/30 x 1/30 = 1/900
but the second part is:
b) a 3 on one card and an 18 on the other card.

The answer to b) is 1/450. I'm a little confused how it is different to the first part and how they got the different answer!?

Thanks

[fun_jirachi]

Hello!
Could someone please help me with the following question:
In a set of 30 cards, each one has a number on it from 1 to 30. If 1 card is drawn out, then replaced and another drawn out, find the probability of getting:
a) a 3 on the first card and an 18 on the second card (I have this answer --> 1/30 x 1/30 = 1/900
but the second part is:
b) a 3 on one card and an 18 on the other card.

The answer to b) is 1/450. I'm a little confused how it is different to the first part and how they got the different answer!?

Thanks

ok the answer is correct
this is because 3 on one card and 3 on another means they can be in any order ie. 3 first, 18 second or 18 first 3 second.
part a) you had the specific scenario of 3 first 18 second, so the answer was 1/900, but since you have a second identical scenario, its 2/900, or 1/450

[Mate2425]

Hey guys how would i approach this question. Names of 10 boys and 8 girls are placed in a hat. Another set of different names of 8 boys and 5 girls are placed in another hat. One name is selected from each hat.

i) What is the probability that the name of a girl and then a boy is picked?
ii) What is the probability of getting a boy and a girl?

Thank you.

[Fergus6748]

Hey guys how would i approach this question. Names of 10 boys and 8 girls are placed in a hat. Another set of different names of 8 boys and 5 girls are placed in another hat. One name is selected from each hat.

i) What is the probability that the name of a girl and then a boy is picked?
ii) What is the probability of getting a boy and a girl?

Thank you.

Heya, honestly the easiest way to approach this would be to make a tree diagram with two events. So the first branches is the first hat, while the next section is the second hat. So:
i) Probability of event 1 being a girl times the proability of event 2 being a boy
(8/18) x (8/13) = 64/234
ii) Simply, the above answer times two. If you draw out the tree diagram there are four possibilites; Boy/Boy, Boy/Girl, Girl/Boy, and Girl/Girl
(64/234) + (64/234) = 128/234

[fun_jirachi]

ii) Simply, the above answer times two. If you draw out the tree diagram there are four possibilites; Boy/Boy, Boy/Girl, Girl/Boy, and Girl/Girl
(64/234) + (64/234) = 128/234

isnt the answer a boy in the first girl in the second + girl in the first and boy in the second
im pretty sure, correct me if im wrong, B1G2 doesnt equal G1B2
so the answer is (8/18) x (8/13) + (10/18) x (5/13) = 114/234 = 19/39

[Fergus6748]

isnt the answer a boy in the first girl in the second + girl in the first and boy in the second
im pretty sure, correct me if im wrong, B1G2 doesnt equal G1B2
so the answer is (8/18) x (8/13) + (10/18) x (5/13) = 114/234 = 19/39

Oh right, yeah you're right, I just assumed they were the same

[amelia20181]

how do you do

[fun_jirachi]

you find where y=1-x^2 cuts the x-axis
y=1-x^2
  =(1+x)(1-x)
ie. it cuts the x-axis at -1 and 1
as such we integrate y=1-x^2 between 1 and -1, or we can just integrate y=1-x^2 from 1 to 0 and multiply the result by two, seeing as it is an even function
integrating 1-x^2 between 1 and 0 gets you 2 x ((1-(1^3)/3)-(0-(0^3)/3))
= 2 x 2/3 = 4/3

[amelia20181]

how do you do

[fun_jirachi]

answer to the above