Answer:
$372,000
Step-by-step explanation:
So, first I'm going to find the equation.
The basic linear equation is y=mx+b
We know there is a difference of 5 years between 155,000 and 310,000. Subtract 155,000 from 310,000 and we get 155,000.
Divide 155,000 by 5, and we get 31,000. It increases by 31,000 every year.
Now we have y=31,000x+b.
To find b we can plug stuff in. y=sales and x=year. Im going to use the third year results.
155,000=31,000(3)+b
155,000=93,000+b
-93,000 -93,000
62,000=b
Now we have our equation y=31,000x+62,000.
Plus in 10 as x to find year 10"s sales and.... we get $372,000
The tenth year sales is $372000
We will use arithmetic progression to solve this. The formula for the nth term is a + (n-1)d.
In their third business year, sales were $155,000. This can be represented as:
a + 2d = 155000 ......... equation i
In the eighth business year, sales were $310,000 and this will be:
a + 7d = 310000 ......... equation ii
a + 2d = 155000 ...... i
a + 7d = 310000 ....... ii
Subtract equation i from ii
(7d - 2d) = (310000 - 155000)
5d = 155000
Divide both side by 5
5d/5 = 155000/5
d= 31000
Then, we can get the value of a which is 1st year sales and this will be:
a + 2d = 155000
a + 2(31000) = 155000
a + 62000 = 155000
a = 155000 - 62000
a = 93000
Then, the 10th year sales will be:
= a + (n - 1)d
= a + (10 - 1)d
= a + 9d
= 93000 + 9(31000)
= 93000 + 279000
= $372000
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