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.

The formula for simple interest is:

I = Prt, where I= the amount of interest accumulated, P = the principal (the initial amount invested), r = the interest rate, and t = the amount of time the money accrues interest

In our problem, P = 300, A= 6% = .06, and t = 6 (years)

So our equation becomes:

I = 300(.04)(6)

Once you perform the above calculation, Darrick, add this amount to the initial deposit of $300 to obtain the balance of the account after 6 years.

Step-by-step explanation: