A student looking at global population figures remarks that between the years 1950 and 2010 the global population can be modelled by the equation 300y= 23t−44 180 (1950≤t≤2010), Write down the gradient of the straight line represented by the equation 300y= 23t−44 180 (1950≤t≤2010)

What does this represent in the practical situation being modelled?

The equation is (i guess):

300*y = 23*t  -41,180

This is a linear equation, where y represents the population, and t represents time (in years)

Now, to find the gradient, we must derive the function y with respect to the time.

then we can write:

y = (23/300)*t - (41,180)/300.

now, remember that when we have:

y = a*x^n

we also have:

dy/dx = n*a*x^(n-1)

then in this case, we have:

dy/dt = (23/300) + 0

now we pass aain the 300 to the right:

300*dy/dt = 23.

d(300*y)/dt = 23

what does this mean?

Thie mean that for each unit of that t increases, we have an increase of 23 units in 300*y.

x = 12

y = 1

for the points!

answer: 5 17/33 % (to the tenth)

step-by-step explanation:

he guessed: 34.5 pounds

actual weight is 32.7

there for he is off:

34.5 -32.7 = 1.8

32.7 is the 100%

1.8 represents %? of 32.7

therefore:

32.7   - 100%

1.8     -     x%

x= 1.8*100%/32.7% = 18*100%/327%

x=1800%/327%   = (5 165/327)%

to the tenth:

5 17/33 %