Question

this question has two parts. first, answer part a. then, answer part b.
part a
hiking tim began a mountain hike near big bear lake, california at 9:00 a.m. by 10:30 a.m., his elevation is 7200 feet above sea level. at 11:15 a.m., he is at an elevation of 7425 feet above sea level.
a. define variables to represent the situation.
let select choice be his altitude above sea level in feet and let select choice be the number of hours tim has been hiking.
part b
b. write an equation in slope - intercept form that represent tim’s elevation since he began hiking.
y = x -
c. if tim’s altitude continues to increase at the same rate, estimate his altitude at 12:30 p.m.
feet above sea level.

Explanation:

Part A

Step 1: Define Variables

Let \( y \) be Tim's altitude above sea level (in feet) and \( x \) be the number of hours Tim has been hiking.

Part B

Step 1: Calculate Slope

First, find the time intervals. From 9:00 a.m. to 10:30 a.m., the time elapsed is \( 1.5 \) hours (since 10:30 - 9:00 = 1 hour 30 minutes = \( 1.5 \) hours), and the altitude change is \( 7200 \) feet (assuming he started at 0? Wait, no, wait. Wait, at 9:00 a.m., let's assume that's the start time, so \( x = 0 \) at 9:00 a.m. Then at 10:30 a.m. (\( x = 1.5 \)), \( y = 7200 \)? Wait, no, the problem says "By 10:30 a.m., his elevation is 7200 feet above sea level. At 11:15 a.m., he is at an elevation of 7425 feet above sea level." Wait, so from 10:30 a.m. to 11:15 a.m., the time elapsed is 45 minutes, which is \( 0.75 \) hours. The altitude change is \( 7425 - 7200 = 225 \) feet. So the slope (rate of change) is \( \frac{225}{0.75} = 300 \) feet per hour. Wait, or maybe from 9:00 a.m. to 10:30 a.m. is \( 1.5 \) hours, and 9:00 a.m. to 11:15 a.m. is \( 2.25 \) hours. Let's check both intervals. From 9:00 a.m. (x=0) to 10:30 a.m. (x=1.5), y=7200. From 9:00 a.m. to 11:15 a.m. (x=2.25), y=7425. Then the slope is \( \frac{7425 - 7200}{2.25 - 1.5} = \frac{225}{0.75} = 300 \). Or from x=0 (9:00 a.m.) to x=1.5 (10:30 a.m.), if we assume he started at some initial altitude? Wait, maybe the problem is that at 9:00 a.m., let's say x=0, and his altitude at x=0 is, wait, the problem says "Tim began a mountain hike near Big Bear Lake, California at 9:00 a.m." So maybe at 9:00 a.m., his altitude is, let's see, at 10:30 a.m. (1.5 hours later), he's at 7200, and at 11:15 a.m. (2.25 hours later), he's at 7425. So the slope (m) is \( \frac{7425 - 7200}{2.25 - 1.5} = \frac{225}{0.75} = 300 \) feet per hour. Now, let's find the y-intercept (b). Using the point (1.5, 7200): \( y = mx + b \), so \( 7200 = 300(1.5) + b \). Calculate \( 300(1.5) = 450 \), so \( b = 7200 - 450 = 6750 \). Wait, or if we use x=0, what's y? Wait, maybe I made a mistake. Wait, maybe at 9:00 a.m. (x=0), his altitude is, let's see, from 9:00 a.m. to 10:30 a.m. is 1.5 hours, and he's at 7200. Then from 10:30 a.m. to 11:15 a.m. is 0.75 hours, and he gains 225 feet, so rate is 300 feet per hour. So the equation is \( y = 300x + b \). When x=1.5 (10:30 a.m.), y=7200: \( 7200 = 300(1.5) + b \) → \( 7200 = 450 + b \) → \( b = 6750 \). So the equation is \( y = 300x + 6750 \)? Wait, but that would mean at x=0 (9:00 a.m.), his altitude is 6750 feet? Maybe the problem is that "began a mountain hike" near Big Bear Lake, which is already at some altitude. So that makes sense.

Step 2: Write the Equation (Part Bb)

So the slope-intercept form is \( y = 300x + 6750 \), where \( x \) is the number of hours since 9:00 a.m., and \( y \) is the altitude in feet.

Step 3: Estimate Altitude at 12:30 p.m. (Part Bc)

12:30 p.m. is 3.5 hours after 9:00 a.m. (since 12:30 - 9:00 = 3.5 hours). So substitute \( x = 3.5 \) into the equation: \( y = 300(3.5) + 6750 = 1050 + 6750 = 7800 \) feet. Wait, but let's check the time intervals again. Wait, from 9:00 a.m. to 10:30 a.m. (1.5 hours): x=1.5, y=7200. From 10:30 a.m. to 11:15 a.m. (0.75 hours): x=2.25, y=7425. From 11:15 a.m. to 12:30 p.m. is 1 hour 15 minutes, which is 1.25 hours. So total x from 9:00 a.m. to 12:30 p.m. is 3.5 hours. So using the equation \( y = 300x + 6750 \), x=3.5: \( y = 300*3.5 + 6750 = 1050 + 6750 = 7800 \) feet.

Answer:

s:

Part A

Let \( y \) = Tim's altitude above sea level (in feet), \( x \) = number of hours since 9:00 a.m.

Part B

b. Equation: \( y = 300x + 6750 \)

c. Altitude at 12:30 p.m.: 7800 feet above sea level.