Answered

In 1980, about 2,180,000 us workers worked at home. During the next ten years the number of workers working at home increased 5% per year. Based on this information how many Americans were working at home in 1999? Round to the nearest million



Answer :

zarahaider4211

Answer:

5,508,751

Step-by-step explanation:

Exponential Equations

An exponential equation can model word problems where rapid growth or decay occurs.

                                                  [tex]y=a(b)^x[/tex],

where

  • a is the initial amount,
  • b is the rate of growth (b > 1) or decay (0 < b < 1).

[tex]\hrulefill[/tex]

Solving the Problem

We're told

  • In 1980, there were 2,180,000 workers at home
  • the number of workers increases by 5% per year

and we need to find the number of at-home workers in 1999.

[tex]\dotfill[/tex]

Identifying the Components of Our Exponential Equation

X - Years

Since time is an independent variable like x, our x in our exponential equation can be in years. Specifically, the number of years since 1980.

A - 2,180,000

This means that at x = 0, or 1980, our function's value is 2,180,000:

                                             [tex]2180000=a(b)^0[/tex].

Since any base raised to the zeroth power is 1, our a value is 2,180,000.

                                            [tex]y=2180000(b)^x[/tex]

B - 1.05

The problem says the number of workers grows, so our b value will be 100% + 5% or 1 plus the decimal form of 5% or 1.05. This is because each following year from 1980, the count of workers has to be bigger than the previous year's.

We can also test this out by finding the values in 1981 and 1982.

1981; x = 1:

                          [tex]y=2180000(1.05)^1=2289000[/tex].

If we manually find the value of 5% more than 2180000 then we also get the value

    2180000 + 5% of 2180000 = 2180000 + 109000 = 2289000.

1982; x = 2:

                           [tex]y=2180000(1.05)^2=2403450[/tex].

Manually finding the value of 5% more than the value in 1981 (2289000), we also get the value

    2289000 + 5% of 2289000 = 2289000 + 114450 = 2403450.

[tex]\dotfill[/tex]

Putting it Together

So, our new model

                                         [tex]y=2180000(1.05)^x[/tex]

checks out.

Let's find the value in 1999, which is 19 years after 1980.

                            [tex]y=2180000(1.05)^{19}=5508751[/tex]