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%)
Answer: B
Step-by-step explanation:
Answer:
(p+13)(p+1)
Step-by-step explanation:
p² + 14p + 13
we want to find two numbers that add to 14 and multiply to 13
we can start looking by listing the factors of 13
we get 13 and 1 and -13 and -1
13 and 1 multiply to 13 ( 13 × 1 = 13 ) and add to 14 ( 13 + 1 = 14 )
-13 and -1 multiply to 13 ( -13 × -1 = 13 ) and add to -14 ( -13 + -1 = -14 )
the two numbers that would add to 14 and multiply to 13 would be 13 and 1
now we can factor out the two numbers from 14 and split the p from p² to get (p+13)(p+1)
and we are done
Answer:
12.5
Step-by-step explanation:
Median is the middle number. First put them in order.
12, 12, 12, 13, 14, 15
The middle is between 12 and 13. Add the numbers together and divide by 2.
25/2 = 12.5
I believe they are 30 years old
