Total world consumption of iron was approximately 900 million tons in 1993. If consumption had increased by 2.5% each year and the resources available for mining in 1993 were 450 billion tons, how much longer would the world’s iron resources have lasted?

Step-by-step explanation:

We can model the annual consumption of iron by the equation

c = 900 million tons x , where

n is the number of years after 1993 and

c is the consumption for that year.

An estimate for the total consumption of iron can now be obtained by integrating this equation with respect to a dummy variable, x from 0 to n.

This equation simplifies to

C = 900 million tons x (1.025^n - 1) / ln(1.025), where

x is a dummy variable and

C is the total consumption of iron n years after 1993.

Let R be the amount of iron remaining at n years after 1993, so

R = 480 billion tons - C

R = 480 billion tons - 900 million tons x (1.025^n - 1) / ln(1.025)

We need to find the value of n when R is 0, so

0 = 480 billion tons - 900 million tons x (1.025^n - 1) / ln(1.025)

Simplify this equation to obtain

n = 107.4

So the iron will run out in the year 1993 + 107 = 2100

answer: c, the green line

step-by-step explanation:

there is no black line, eliminate b.

simple interest implies periodic payments, but the red line shows no payments, eliminate d.

simple interest has equal payments. if paid say every day, the graph is about a straight line (actually a stairstep).

the yellow line is not straight, eliminate a.

check: the green line is straight, answer is c.

simple interest with positive interest rate gives a straight line sloping up to the right. green line:

g(t) = $100 + ($200-$100) ÷ (20 years) × (t years)

= $100 + (4%/year) × ($100)(t years)

incidentally, the yellow line curving upward represents continuously compounded interest

y = $100 × 2^(t years / (14 assuming the yellow line hits $200 at 14 years. 2^0 is 1, so y(0) = $100; 2^(14/14) is 2, so y(14) = $200.

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step-by-step explanation: