<span>A ski lift is designed to hold 20,000 pounds, and claims a capacity of 100 persons. suppose the weights of all people using the lift have a mean of 190 pounds and with a standard deviation of 45 pounds. what is the probability that a random group of 100 people will total more than the weight limit of 20,000 pounds?

Given data:
Capacity, x=20,000 lbs
n=100
μ=190
σ=45
&se;=σ/sqrt(n)=45/sqrt(100)=4.5
s=population mean
population μ and σ are both known, use SE normally distributed.
P(s≤x/n)=P(s≤200)=Z((200-190)/4.5)=Z(2.222)=0.98687
=>
P(s≥x/n)=1-P(s≤x/n)=1-0.98687=0.01313

Answer: probability of 100 people exceeding 20,000 lbs is 1.313%</span>

4 0

4 years ago

Which expression is equivalent to one over five m − 20? (4 points)

exis [7]

Answer:

The equivalent to one over five m − 20 is one over five (m − 100) ⇒ B

Step-by-step explanation:

Let us solve the question

∵ One over five means $\frac{1}{5}$

∴ One over five m - 20 = $\frac{1}{5}$ m - 20

→ By using the distributive property, take one over five as a common factor

from both terms

∴ $\frac{1}{5}$ m - 20 = $\frac{1}{5}$ $(\frac{\frac{1}{5}m}{\frac{1}{5}}$ - $\frac{20}{\frac{1}{5}})$

→ Simplify the bracket

∵ $\frac{1}{5}$ m ÷ $\frac{1}{5}$ = $\frac{1}{5}$ m × 5 = m

∵ 20 ÷ $\frac{1}{5}$ = 20 × 5 = 100

∴ $\frac{1}{5}$ $(\frac{\frac{1}{5}m}{\frac{1}{5}}$ - $\frac{20}{\frac{1}{5}})$ = $\frac{1}{5}$ (m - 100)

∴ $\frac{1}{5}$ m - 20 = $\frac{1}{5}$ (m - 100)

The equivalent to one over five m − 20 is one over five (m − 100)

4 0

4 years ago