Which criteria are responsible for deciding whether a heterogeneous mixture Is a colloid or a suspension?
2 answers:
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Answer:
A) Enterprise: E(x) = 50 +0.25x; Hertz: H(x) = 30 +0.45x
B) see attached graph
Step-by-step explanation:
A) We choose to compare Hertz (lowest daily charge) and Enterprise (lowest mileage charge). In each case, the total charge will be the sum of the daily charge and the product of the mileage charge and number of miles (x).
Enterprise: E(x) = 50 +0.25x
Hertz: H(x) = 30 +0.45x
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
- PY(k) = PX(k) = ε^(-3)*(3^k)/k! for k in {0,1,2,3}
- PY(4) = 1 -PX(0) - PX(1) - PX(2) - PX (3)
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!