Hey there!
Let's first define our goal. Our goal here is to be able to isolate the variable, n, and solve for it by moving everything to the other side. We see that we have two sides of the equation, and by using the property of equality which states that whatever you do to one side you do to the other and the expression remains equal, we can solve this equation.
It looks like the only thing we have on the left side with n is that 6 1/2. In order to get rid of it, we must subtract it from both sides, because the property of equality states you must do the inverse operation.
We have:
n = 12 - 6 1/2
n = 5 1/2
Hope this helps!
(Odds) I have to put 20 characters
Answer:
60cm
Step-by-step explanation:
total length of wood = 3m
number of sections = 5
width of each section = 3/5 = 0.6 m
= 60cm
Answer:
S = {0,2,3,4}
P(X=0) = 0.573 , P(X=2) = 0.401 , P(x=3) = 0.025, P(X=4) = 0.001
Mean = 0.879
Standard Deviation = 1.033
Step-by-step explanation:
Let the number of people having same birth month be = x
The number of ways of distributing the birthdays of the 4 men = (12*12*12*12)
The number of ways of distributing their birthdays = 12⁴
The sample space, S = { 0,2,3,4} (since 1 person cannot share birthday with himself)
P(X = 0) = 
P(X=0) = 0.573
P(X=2) = P(2 months are common) P(1 month is common, 1 month is not common)
P(X=2) = 
P(X=2) = 0.401
P(X=3) = 
P(x=3) = 0.025
P(X=4) = 
P(X=4) = 0.001
Mean, 

Standard deviation, ![SD = \sqrt{\sum x^{2} P(x) - \mu^{2}} \\SD =\sqrt{ [ (0^{2} * 0.573) + (2^{2} * 0.401) + (3^{2} * 0.025) + (4^{2} * 0.001)] - 0.879^{2}}](https://tex.z-dn.net/?f=SD%20%3D%20%5Csqrt%7B%5Csum%20x%5E%7B2%7D%20P%28x%29%20-%20%5Cmu%5E%7B2%7D%7D%20%20%5C%5CSD%20%3D%5Csqrt%7B%20%5B%20%280%5E%7B2%7D%20%2A%200.573%29%20%2B%20%282%5E%7B2%7D%20%20%2A%200.401%29%20%2B%20%283%5E%7B2%7D%20%2A%200.025%29%20%2B%20%284%5E%7B2%7D%20%2A%200.001%29%5D%20-%200.879%5E%7B2%7D%7D)
SD = 1.033
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