Find an equation for the linear model of the situation below and use it to make a prediction. A train is traveling north at a constant rate. At 3:00 P.M. it is 55 miles north of a city. At 4:15 P.M. it is 80 miles north of the city. If d represents the distance in miles, and t represents the time in hours, how many miles north of the city will the train be at 5:45 P.M.?
„h d = 64t + 55; d = 231 miles
„h d = 80t + 55; d = 255 miles
„h d = 64t + 55; d = 96 miles
„h d = 100t + 55; d = 285 miles
I know the answer is the first choice but I don't understand how to set the problem up.
Well, let's break it down. We are looking for an equation that represents the relationship between the distance the train is from the city and the time it has been traveling.
Let's start with the given information. At 3:00 P.M., the train is 55 miles north of the city. We can use this to determine the y-intercept of our equation.
Next, at 4:15 P.M., the train is 80 miles north of the city. This gives us another point on the line, which we can use to determine the slope of the line.
To find the slope, we need to calculate the change in distance over the change in time. The time difference between 3:00 P.M. and 4:15 P.M. is 1 hour and 15 minutes or 1.25 hours. The distance difference is 80 - 55 = 25 miles.
So, the slope of the line is 25/1.25 = 20.
Now we can plug the slope (m = 20) and the y-intercept (b = 55) into the equation of a line, y = mx + b, to find the equation for the linear model.
Therefore, the equation for the linear model is d = 20t + 55.
To predict the distance the train will be at 5:45 P.M., we substitute t = 2.75 (since 5:45 P.M. is 2 hours and 45 minutes after 3:00 P.M.) into the equation:
d = 20 * 2.75 + 55 = 55 + 55 = 110.
So, according to the linear model, the train will be 110 miles north of the city at 5:45 P.M.
But hey, no matter how far the train is, just remember that it's always on the right track!
To set up the problem, we need to find the equation for the linear model using the given information.
We are given that at 3:00 P.M., the train is 55 miles north of the city. This can be represented as a point on the graph (t, d) = (0, 55), where t is the time in hours and d is the distance in miles.
Next, at 4:15 P.M., the train is 80 miles north of the city. This can be represented as another point on the graph (t, d) = (1.25, 80).
We can use these two points to find the equation of the line using the slope-intercept form (y = mx + b), where m is the slope and b is the y-intercept.
The slope can be found using the formula: m = (y2 - y1) / (x2 - x1)
Substituting the values from the two points, we get:
m = (80 - 55) / (1.25 - 0) = 25 / 1.25 = 20
Now, we have the slope (m = 20). We can substitute this value along with one of the points (0, 55) into the equation y = mx + b to find the y-intercept (b).
55 = 20(0) + b
55 = b
So, the equation for the linear model is d = 20t + 55.
To find out how many miles north of the city the train will be at 5:45 P.M., we need to substitute t = 2.75 into the equation:
d = 20(2.75) + 55 = 55 + 55 = 110
Therefore, the train will be 110 miles north of the city at 5:45 P.M.
To find the equation for the linear model, we need to determine the relationship between the distance traveled by the train and the time elapsed.
First, we can identify two data points from the given information:
- At 3:00 P.M., the train is 55 miles north of the city.
- At 4:15 P.M., the train is 80 miles north of the city.
Next, we can calculate the time elapsed between these two points.
From 3:00 P.M. to 4:15 P.M., the time elapsed is 1 hour and 15 minutes, which is equivalent to 1.25 hours.
Now, we can set up a linear equation in the form of "d = mt + b," where d represents the distance and t represents the time. We need to find the slope (m) and the y-intercept (b) to complete the equation.
Using the two data points:
(3:00 P.M., 55 miles)
(4:15 P.M., 80 miles)
We can find the slope (m):
m = (change in distance) / (change in time)
m = (80 - 55) / (1.25)
m = 25 / 1.25
m = 20
Now, we can substitute the slope (m) into the equation:
d = 20t + b
To find the y-intercept (b), we can substitute one of our data points into the equation:
55 = 20(t) + b
55 = 20(3) + b
55 = 60 + b
b = -5
Therefore, the equation for the linear model is:
d = 20t - 5
To predict the distance north of the city at 5:45 P.M. (which is 2 hours and 45 minutes past 3:00 P.M.), we substitute t = 2.75 into the equation:
d = 20(2.75) - 5
d = 55 + 45 - 5
d = 95
Hence, the train will be 95 miles north of the city at 5:45 P.M.