Answer:
x=15
Step-by-step explanation:
subtract 1 from both sides so you are left with 2x=5x-45
then subtract 5x from both sides and you have -3x=-45
then finally divide -3 from -45 to get x=15
Answer:

In order to find the variance we need to find first the second moment given by:

And replacing we got:

The variance is calculated with this formula:
![Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075](https://tex.z-dn.net/?f=%20Var%28X%29%20%3D%20E%28X%5E2%29%20-%5BE%28X%29%5D%5E2%20%3D%200.33%20-%280.15%29%5E2%20%3D%200.3075)
And the standard deviation is just the square root of the variance and we got:

Step-by-step explanation:
Previous concepts
The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.
The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).
Solution to the problem
LEt X the random variable who represent the number of defective transistors. For this case we have the following probability distribution for X
X 0 1 2 3
P(X) 0.92 0.03 0.03 0.02
We can calculate the expected value with the following formula:

And replacing we got:

In order to find the variance we need to find first the second moment given by:

And replacing we got:

The variance is calculated with this formula:
![Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075](https://tex.z-dn.net/?f=%20Var%28X%29%20%3D%20E%28X%5E2%29%20-%5BE%28X%29%5D%5E2%20%3D%200.33%20-%280.15%29%5E2%20%3D%200.3075)
And the standard deviation is just the square root of the variance and we got:

Answer:
t = 13.1576 years
Step-by-step explanation:
The situation can be modeled as
P = Po.(1-r)^t
Where
t is the years transcurred
Po is the initial amount
r is the rate of change
Po = 40000
r = 10% equivalent to 0.1
Now
A quarter of the original amount
(1/4)*Po = Po.(1-r)^t
(1/4) = (0.9)^t
t = 13.1576
Please, see attached picture
Answer:
The probability of 1 or less children from that group to learn how to swim before 6 years of age is 0.072
Step-by-step explanation:
In this case we need to compute the probability of none of these 12 children learns to swim before 6 years of age. This is given by:
p(0) = (1 - 0.312)^(12) = 0.688^(12) = 0.01124
We now need to calculate the probability that one child learns to swim before 6 years of age.
p(1) = 12*0.312*(1 - 0.312)^(11) = 3.744*(0.688)^(11)
p(1) = 3.744*0.01634
p(1) = 0.0612
The probability of 1 or less children from that group to learn how to swim before 6 years of age is:
p = p(0) + p(1) = 0.01124 + 0.0612 = 0.07244