Each pound of potato salad serves 6 people and the average number of pounds of potato salad that each person is served will be 
pound.
<u><em>Explanation</em></u>
At a party of 102 people, 17 pounds of potato salad is served.
<u>a.</u> Suppose, each pound of potato salad is served to 
number of people.
So, 17 pounds of salad is served to 
or 
number of people. That means......
So, each pound of potato salad is served to 6 people.
<u>b.</u> Suppose, the average number of pounds of potato salad that each person is served is 
So, the total number of pounds of salad, served to 102 people 
pounds 
pounds. That means .....
So, the average number of pounds of potato salad that each person is served will be pound.
Part A: r = d/t
Part B: 221760 feet per hour
Part C: 42 mph
Part D: Miles because we are measuring large distances
Answer:
The proportion of students whose height are lower than Darnell's height is 71.57%
Step-by-step explanation:
The complete question is:
A set of middle school student heights are normally distributed with a mean of 150 centimeters and a standard deviation of 20 centimeters. Darnel is a middle school student with a height of 161.4cm.
What proportion of proportion of students height are lower than Darnell's height.
Answer:
We first calculate the z-score corresponding to Darnell's height using:
We substitute x=161.4 , 
, and 
to get:
From the normal distribution table, we read 0.5 under 7.
The corresponding area is 0.7157
Therefore the proportion of students whose height are lower than Darnell's height is 71.57%
Answer:
g(1) = 0
=
lgn= ) +m
g(n) = g(n - 1) +n
=
-
n
g(2) =12
Step-by-step explanation:
12345678910
Might have to experiment a bit to choose the right answer.
In A, the first term is 456 and the common difference is 10. Each time we have a new term, the next one is the same except that 10 is added.
Suppose n were 1000. Then we'd have 456 + (1000)(10) = 10456
In B, the first term is 5 and the common ratio is 3. From 5 we get 15 by mult. 5 by 3. Similarly, from 135 we get 405 by mult. 135 by 3. This is a geom. series with first term 5 and common ratio 3. a_n = a_0*(3)^(n-1).
So if n were to reach 1000, the 1000th term would be 5*3^999, which is a very large number, certainly more than the 10456 you'd reach in A, above.
Can you now examine C and D in the same manner, and then choose the greatest final value? Safe to continue using n = 1000.
