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.
We would need a sample size of 560.
We first calculate the z-score associated. with this level of confidence:
Convert 95% to a decimal: 95% = 95/100 = 0.95
Subtract from 1: 1-0.95 = 0.05
Divide by 2: 0.05/2 = 0.025
Subtract from 1: 1-0.025 = 0.975
Using a z-table (http://www.z-table.com) we see that this is associated with a z-score of 1.96.
The margin of error, ME, is given by:

We want ME to be 4%; 4% = 4/100 = 0.04. Substituting this into our equation, as well as our proportion and z-score,
It's C.
multiply by 3
x = 4, y = 4x3 = 12
x = 5, y = 5x3 =15
x = 6, y = 6x3 =18
Answer:
x^2+5x-14
Step-by-step explanation:
you multiply both equations so that you get:
x^2-2x+7x-14
and then you simplify it to:
x^2+5x-14
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<em><u>I would also appreciate it if you give me a brainliest</u></em>