Number of students = 9860
Total Population = 62,400
Percentage of students in the population can be calculated by:
(Number of students/Total population) x 100%
Using the values, we get:
Percentage of students = (9860/62400) x 100% = 15.80%
Thus, students constitute 15.80% of the entire population
Answer:
6 knots
Step-by-step explanation:
Let the speed be v knots
then time taken to cover 500 M = 500 / v hrs
fuel consumption /hr = 216 + 0.5v^3
let F be the fuel consumption for trip
= [500/v][216 + 0.5v^3]
= 500[216/v + 0.5v^2]
dF/dv = 500[ - 216/v^2 + v]
d^2F/d^2v = 500[432/v^3 + 1] , i.e. +ve
so setting dF/dv will give a minima
500[ -216/v^2 + v] = 0
or v = 216/v^2
or v^3 = 216
solving, we get v = [216]^(1/3) = 6 knots
Rates like $ per channel is a slope, "m". The added fee is a constant so it's the intercept "b".
y = mx + b
So for the first problem (9)
(a)
y = total cost in dollars
x = number of premium channels
y = 16x + 44
(b) when x = 3 channels
y = 16(3) + 44
y = 92 $
the second problem (10)
(a) every 4 years the tree grows by 12-9=3 ft
So the unit rate or slope will be 3 ft per 4 yrs, (3/4). You can see this also by solving for slope "m" using the given points (4,9) and (8,12).
x = number of years
y = height of tree in ft
y = (3/4)x + b
use one of the points to find the y-intercept "b".
9 = (3/4)(4) + b
9 = 3 + b
9 - 3 = b
6 = b
y = (3/4)x + 6
(b) when x = 16
y = (3/4)(16) + 6
y = 12 + 6
y = 18 ft
Answer:
<h2><em>
30page essay/min</em></h2>
Step-by-step explanation:
Speed is the change in distance of a body with respect to time.
Speed = Distance/Time
If the library made a copy of leslie's 3 page essay in just 1/10 of a minute, this means that she made 3 pages in (1/10 * 60)secs i.e 6secs
To get the speed, we need to calculate the amount of page of essay that are produced in a minute (60secs).
If 3 page essay = 6secs
x page essay = 60 secs
cross multiply
3*60 = 6x
180 = 6x
x = 180/6
x = 30
<em>This shows that 30 page essays are produced in just a minute. Hence the speed of the copy machine is 30page essay/min.</em>
