You would have ran 1,200 meters
1/5 and 3/5
Hope this helps!
Let d represent the distance of the destination from the starting point.
After 45 min, Henry has already driven d-68 miles. After 71 min., he has already driven d-51.5 miles.
So we have 2 points on a straight line:
(45,d-68) and (71,d-51.5). Let's find the slope of the line thru these 2 points:
d-51.5 - (d-68) 16.5 miles
slope of line = m = ----------------------- = ------------------
71 - 45 26 min
Thus, the slope, m, is m = 0.635 miles/min
The distance to his destination would be d - (0.635 miles/min)(79 min), or
d - 50.135 miles. We don't know how far his destination is from his starting point, so represent that by "d."
After 45 minutes: Henry has d - 68 miles to go;
After 71 minutes, he has d - 51.5 miles to go; and
After 79 minutes, he has d - x miles to go. We need to find x.
Actually, much of this is unnecessary. Assuming that Henry's speed is 0.635 miles/ min, and knowing that there are 8 minutes between 71 and 79 minutes, we can figure that the distance traveled during those 8 minutes is
(0.635 miles/min)(8 min) = 5.08 miles. Subtracting thix from 51.5 miles, we conclude that after 79 minutes, Henry has (51.5-5.08), or 46.42, miles left before he reaches his destination.
Let us model this problem with a polynomial function.
Let x = day number (1,2,3,4, ...)
Let y = number of creatures colled on day x.
Because we have 5 data points, we shall use a 4th order polynomial of the form
y = a₁x⁴ + a₂x³ + a₃x² + a₄x + a₅
Substitute x=1,2, ..., 5 into y(x) to obtain the matrix equation
| 1 1 1 1 1 | | a₁ | | 42 |
| 2⁴ 2³ 2² 2¹ 2⁰ | | a₂ | | 26 |
| 3⁴ 3³ 3² 3¹ 3⁰ | | a₃ | = | 61 |
| 4⁴ 4³ 4² 4¹ 4⁰ | | a₄ | | 65 |
| 5⁴ 5³ 5² 5¹ 5⁰ | | a₅ | | 56 |
When this matrix equation is solved in the calculator, we obtain
a₁ = 4.1667
a₂ = -55.3333
a₃ = 253.3333
a₄ = -451.1667
a₅ = 291.0000
Test the solution.
y(1) = 42
y(2) = 26
y(3) = 61
y(4) = 65
y(5) = 56
The average for 5 days is (42+26+61+65+56)/5 = 50.
If Kathy collected 53 creatures instead of 56 on day 5, the average becomes
(42+26+61+65+53)/5 = 49.4.
Now predict values for days 5,7,8.
y(6) = 152
y(7) = 571
y(8) = 1631
