Answer:Total units would be: 3+4 = 7. value of 1 unit = 28/7 = 4. ratio would be 3(4) & 4(4) = 12 & 16
Step-by-step explanation:
Answer:
For x(θ) = 3cosθ + 2
y² = [9x²/(x - 2)²] - x²
For x(θ) = 3cosθ + 2
x² = [4y²/(y - 1)²] - y²
Step-by-step explanation:
Given the following equivalence:
x² + y² = r²
r = √(x² + y²)
x = rcosθ
cosθ = x/r
y = rsinθ
sinθ = y/r
Applying these to the given equations,
x(θ) = 3cosθ + 2
x = 3(x/r) + 2
xr = 3x + 2r
(x - 2)r = 3x
r = 3x/(x - 2)
Square both sides
r² = 9x²/(x - 2)²
(x - 2)²r² = 9x²
(x - 2)²(x² + y²) = 9x²
(x² + y²) = 9x²/(x - 2)²
y² = [9x²/(x - 2)²] - x²
y(θ) = 2sinθ - 1
y = 2y/r - 1
yr = 2y - r
(y - 1)r = 2y
r = 2y/(y - 1)
Square both sides
r² = 4y²/(y - 1)²
x² + y² = 4y²/(y - 1)²
x² = [4y²/(y - 1)²] - y²
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%)
