Step-by-step explanation: The base of the power in the original equation becomes the base of the log. So we have .
Next, the exponent in the original equation goes on the other side of the equation and finally, the result in the original equation goes inside the log.
So we have
which is 3² = 9 written in logarithmic form.
This can also be switched around to read .
So the first choice will be your answer.
My work is also show in the image attached,
I just didn't switch it.
Answer:
Sqrt 3
Step-by-step explanation:
Answer:
a) The expected value is 2.680642
b) The minimun number of newspapers the manager should order is 6.
Step-by-step explanation:
a) Lets call X the demanded amount of newspapers demanded, and Y the amount of newspapers sold. Note that 4 newspapers are sold when at least four newspaper are demanded, but it can be <em>more</em> than that.
X is a random variable of Poisson distribution with mean , and Y is a random variable with range {0, 1, 2, 3, 4}, with the following values
we obtain:
PY(0) = ε^(-3) = 0.04978..
PY(1) = ε^(-3)*3^1/1! = 3*ε^(-3) = 0.14936
PY(2) = ε^(-3)*3^2/2! = 4.5*ε^(-3) = 0.22404
PY(3) = ε^(-3)*3^3/3! = 4.5*ε^(-3) = 0.22404
PY(4) = 1- (ε^(-3)*(1+3+4.5+4.5)) = 0.352768
E(Y) = 0*PY(0)+1*PY(1)+2*PY(2)+3*PY(3)+4*PY(4) = 0.14936 + 2*0.22404 + 3*0.22404+4*0.352768 = 2.680642
The store is <em>expected</em> to sell 2.680642 newspapers
b) The minimun number can be obtained by applying the cummulative distribution function of X until it reaches a value higher than 0.95. If we order that many newspapers, the probability to have a number of requests not higher than that value is more 0.95, therefore the probability to have more than that amount will be less than 0.05
we know that FX(3) = PX(0)+PX(1)+PX(2)+PX(3) = 0.04978+0.14936+0.22404+0.22404 = 0.647231
FX(4) = FX(3) + PX(4) = 0.647231+ε^(-3)*3^4/4! = 0.815262
FX(5) = 0.815262+ε^(-3)*3^5/5! = 0.91608
FX(6) = 0.91608+ε^(-3)*3^6/6! = 0.966489
So, if we ask for 6 newspapers, the probability of receiving at least 6 calls is 0.966489, and the probability to receive more calls than available newspapers will be less than 0.05.
I hope this helped you!