How do I know which numbers the variables r and s represent?
(1) 2-r=s
(2) 1+2r=4-s
(3) 4+r= 2+3s
Gladly with calculation method.
(1) 2-r=s
(2) 1+2r=4-s
(3) 4+r= 2+3s
Gladly with calculation method.
Where is my mistake, I can't find it… (please ignore the formalities)
The task is: In the Lotto “6 out of 49” the random variable “number of correct numbers per tip” is Probability distribution (rounded) given in the table. Check the information for P(X=0), P(X=1), P(X=6) You don’t need any other probabilities here except: P(X=0)= 0.436 ; P(X=1)= 0.413 ; P(X=6)= 7.15*10^-8 . I already have the…
I urgently need an answer! Thanks!
Unfortunately, I'm really bad at math and tried to solve this problem. Can someone correct it for me? Thank you.
a) I was still able to solve; 6!/(3!*1!*1!*1!)=6!/3!=120 I don't understand b) and c) anymore…
= 2-r
Insert in (2):
1+2r = 4-(2-r)
Forming:
2r = 3-(2-r)
2r = 3-2+r
2r = 1+r
R = 1
Insert in (1):
2-1 =s
= 1
You can make the sample if you use in r and s in any equation.
(1) 2 – r = s
(1) r = 2 – s
(1) 2 * r = 4 – 2 * s
(1) 1 + 2 * r = 5 – 2 * s
(2) 1 + 2 * r = 4 – s
That means:
5 – 2 * s = 4 – s
= 1
Because of r=2 -s, the following follows:
R = 1
Basics from the 7th/8th class must be repeated!
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two unknowns
three glazes
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replace all r (or s) by (1)
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s is replaced :
1 + 2r = 4 – (2-r)
1+2r = 2 + r
2r = 2-1
R = 1
in the United Kingdom
2-1 = = 1
.
Sample with (3)
4 + 1 = 2 + 3*1
5 = 5
I’d put the first equation in the third one. We know: s=2-r. So you can replace in the third equation s by 2-r. Looks like this.
4+r= 2+ 3(2-r) then on each side -2, then looks like this:
2+r= 3(2-r) dissolve clamp:
2+r= 6 – 3r
And now I don’t have a bat. I’m only 70% sure if that’s the way. You have to keep looking that you have on your own on one side of the equation. And on the other hand only one number. Then you know what’s r and you can use it in the first equation. And then you know what it is. Have fun! If you don’t get it completely, report to Mathe and say where the problem was.