solve this maths problem
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#70 do I hear a normal distribution?
#83 you can’t solve that. If it was like “2 x :) + :( = 19; :) x :( = 42” you can
#85 wait no never mind after a quick google search I haven't learnt that stuff yet, I think it's like year 12 maths extension 1 here in Australia
WHAT STATE DO U LIVE IN
carrotboy1234 wrote:
#85 wait no never mind after a quick google search I haven't learnt that stuff yet, I think it's like year 12 maths extension 1 here in Australia
You don't need any advanced math to do it. I did it just using factorials.
5+8x98+y=9716+726 y=?
nope
Paz_Hobbitt12 wrote:
I need help
How can I reduce this:
3 - 4y + 10x - 4 - 3y - 10y
How can I reduce this:
3 - 4y + 10x - 4 - 3y - 10y
Do grouping
Group the just integers, the ones with variable y, and the ones with variable x
3 - 4y + 10x - 4 - 3y - 10y = 3-4 -4y-3y-10y +10x
3-4 is -1
-4y-3y-10y is -17y
10x remains the same since there is just one term
So final answer will be -1 -17y +10x
Hope it made sense
#89 ah I see I know what factorials are but haven’t formally learnt those yet either. It’s gotta be like (all possible ways there can be 50 heads)/2^100 or something but idk how to work out the numerator. Not sure if you can just represent the numerator as a half of something or if it’s more complex.
x! is the number of ways x number of things can be arranged. 2 things can be arranged in two ways, so 2! = 2. 4 things can be arranged in 24 ways, so 4!=24. A simple formula is
x! = x * (x-1) * (x-2) . . . * 1 (product of all positive integers till x)
Here's a clue on how I solved it, first I figured out the number of ways 50 heads and 50 tails can be arranged, aka 100!, and divided it by a certain number to cancel out the repetitions. "The certain number" was probably the hardest thing for me to find, but give it a shot.
I finished algebra this year. I didn’t take exams cuz I was exempt so I basically forgot everything? But try testing me