36 all together.
22 first
14 second
8 random chosen
A) all first shift:
One is pulled 22/36
Second is pulled 21/35
Third is pulled 20/34
Fourth 19/33
Fifth 18/32
Sixth 17/31
Seventh 16/30
Eighth 15/29
Multiply all those together
Probability of all first shift is 0.010567296996663
(That means it's not happening anytime soon lol)
B) one worker 14/36
Second 13/35
Third 12/34
Fourth 11/33
Fifth 10/32
Sixth 9/31
Seventh 8/30
Eighth 7/29
Multiply all those together
Probability of all second shift is 0.000099238805645
(That means it's likely to see 100x more picks of all first shift workers before you see this once.. lol)
C) 22/36
21/35
20/34
19/33
18/32
17/31
Multiply..
Probability.. 0.038306451612903
D) 14/36
13/35
12/34
11/33
X... p=0.016993464052288
Probably not correct, haven't done probability in years.
Begin solving like you normally would. Add 2.25 to both sides and subtract 9x from both sides to try getting all x values on one side. However, you will find that after subtracting 9x from both sides, all x values go away, and you end with -2.25=1.6. This is not true (-2.25 is not 1.6), so the answer is no solution. There is no value of x which satisfies that equation.
Answer = 5.656854249
Step by step:
This equation is just a simple Pythagorean theorem ( if you were not aware, it’s a2 + b2 = c2. To get c is to find the square root of c)
In order to find what one part is in a ratio, you have to add the ratio up ( 5+ 3) and divide it by the number you're looking for (56). In this case, you get 56/8, which gives you 7. Therefore, each part is worth 7. You then have to multiply both sides of the ratio (5 and 3) by 7. 5x7= 35. 3x7= 21.
Therefore, 56 divided into the ratio of 5:3 is 35:21
No, because something extra is being added/divided and there are too many variables on one side
