A) maximum mean weight of passengers = <span>load limit ÷ number of passengers
</span><span>
maximum mean weight of passengers = 3750 </span>÷ 25 = <span>150lb
</span>B) First, find the z-score:
z = (value - mean) / stdev
= (150 - 199) / 41
= -1.20
We need to find P(z > -1.20) = 1 - P(z < -1.20)
Now, look at a standard normal table to find <span>P(z < -1.20) = 0.11507, therefore:
</span>P(z > -1.20) = 1 - <span>0.11507 = 0.8849
Hence, <span>the probability that the mean weight of 25 randomly selected skiers exceeds 150lb is about 88.5%</span> </span>
C) With only 20 passengers, the new maximum mean weight of passengers = 3750 ÷ 20 = <span>187.5lb
Let's repeat the steps of point B)
z = (187.5 - 199) / 41
= -0.29
P(z > -0.29) = 1 - P(z < -0.29) = 1 - 0.3859 = 0.6141
</span>Hence, <span>the probability that the mean weight of 20 randomly selected skiers exceeds 187.5lb is about 61.4%
D) The mean weight of skiers is 199lb, therefore:
number</span> of passengers = <span>load limit ÷ <span>mean weight of passengers
= 3750 </span></span><span>÷ 199
= 18.8
The new capacity of 20 skiers is safer than 25 skiers, but we cannot consider it safe enough, since the maximum capacity should be of 18 skiers.</span>
The average rate of change is 0.19
Answer:
<h2>x = 5.7</h2>
Step-by-step explanation:
To find x first cross multiply
We have
7x = 8 × 5
7x = 40
Divide both sides by 7
That's
x = 5.7142
We have the final answer as
<h3>x = 5.7 to the nearest tenth</h3>
Hope this helps you
Answer:
12
Step-by-step explanation:
Listing the weeks he will attend each item
Class 4,8,12,16
Chess 2,4,6,8,10,12
Fencing 3,6,9,12
Starting today at week 0, he will attend all 3 again at week 12
Answer:
b. They can both be 90 degree angles
c. Acute angles are by definition 90 degrees, and acute is less than but not equal to 90 degrees
d. should be always
e. should be never (explanation: they have to be adjacent to be supplementary, and vertical angles are never adjacent)
Step-by-step explanation:
