On one of its routes across asia, alpha airlines flies an aircraft with checked-in luggage capacity of 8500 lbs. there are 121 seats on the flight.
the average (per passenger) weight of checked-in luggage is 68 lbs with a standard deviation of 11 lbs.
what is the probability that on a randomly selected full flight on this route the checked-in luggage capacity will be exceeded?

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%)

we know that

applying the law of sines

°

so

∠j=18°

∠k=29°

∠l=180-(18+29)=133°

the answer part a) is

the measure of angle k is  ≈29°

part b)  the measure of angle k could also be

(180-29)=151°

so

∠j=18°

∠k=151°

∠l=180-(18+151)=11°

the answer part b) is

the measure of angle k could also be 151°

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answer: 59

step-by-step explanation: