B is correct because I know
Red and Green are the answers
Answer:
c
Step-by-step explanation:
Answer:
the probability is P=0.012 (1.2%)
Step-by-step explanation:
for the random variable X= weight of checked-in luggage, then if X is approximately normal . then the random variable X₂ = weight of N checked-in luggage = ∑ Xi , distributes normally according to the central limit theorem.
Its expected value will be:
μ₂ = ∑ E(Xi) = N*E(Xi) = 121 seats * 68 lbs/seat = 8228 lbs
for N= 121 seats and E(Xi) = 68 lbs/person* 1 person/seat = 68 lbs/seat
the variance will be
σ₂² = ∑ σ² (Xi)= N*σ²(Xi) → σ₂ = σ *√N = 11 lbs/seat *√121 seats = 121 Lbs
then the standard random variable Z
Z= (X₂- μ₂)/σ₂ =
Zlimit= (8500 Lbs - 8228 lbs)/121 Lbs = 2.248
P(Z > 2.248) = 1- P(Z ≤ 2.248) = 1 - 0.988 = 0.012
P(Z > 2.248)= 0.012
then the probability that on a randomly selected full flight, the checked-in luggage capacity will be exceeded is P(Z > 2.248)= 0.012 (1.2%)
Since a number is divisible by 2 and 3 and 5 it can be written as 2*3*5*x=num where x is a natural number
a number is only divisible by a number if they share the same prime divisors so the numbers that will always be divisible by that number are:2,3,5,2*3:6,2*5:10,3*5:15,2*3*5:30 C)30 which is the correct answer
