An airline experiences a no-show rate of 6%. What is the maximum number of reservations that it could accept for a flight with a capacity of 160, if it wants the probability of accommodating all reservation holders to be at least

Step-by-step explanation:

Use the normal approximation to the binomial distribution

mean µ = np

standard deviation σ = √npq

Where,

n is sample size

p is probability of success.

q is probability of failure

Given that

q = 6% =0.06

Then, p = 1-q = 1-0.06 = 0.94

Therefore:

µ = pn

µ = 0.94n

Also

σ = √npq

σ = √(n)(0.94)(0.06)

σ = √(.0564n)

Using z-scores:

z = (x — µ )/σ

Using the data above

1.645 = (160 — 0.94n)/√(0.0564n)

Cross multiply

1.645√0.0564n = 160—0.94n

Square both sides

1.645²× 0.0564n = (160-0.94n)

0.153n=25600— 300.8n + 0.8836n²

0.8836n²-300.8n-0.153n +25600=0

0.8836n² — 300.953n + 25600 = 0

Using quadratic formula method.

a = 0.8836 b = -300.953 c = 25600

n = [-b±√(b²-4ac)]/2a

n = [--300.953±√((-300.953)²-4×0.8836×25600)] / (2 × 0.8836)

n = [300.953±√(92.07)]/1.7672

n = (300.953±9.6)/1.762

n = (300.953-9.6)/1.762

n = 168.22

Or

n = (300953+9.6)/1.762

n = 176.25

The maximum number of reservation is approximately 168.

the answer is: d) o, f, and kr.

atomic radius of bromine (br) is 114 pm.

atomic radius of oxygen (o) is 73 pm.

atomic radius of fluorine (f) is 72 pm.

atomic radius of krypron (kr) is 112 pm.

the atomic radius of a chemical element is a measure of the size of its atom.

the atomic radius varies with increasing atomic number, but usually increases because of increasing of number of electrons.

the atomic radius decreases across the periods because an increasing number of protons, because greater attraction between the protons and electrons.

after going and makeing graphs i would go with a,

there is nothing that is matches the graph. if there was i would show you. but therefore my answer is a.

hope i

-lavira: )