Question

1. in most cases, temperature decreases with increasing altitude. as anchara drives into the mountains, her car thermometer registers the temperatures (f) shown in the table at the given altitudes (feet).

altitude | 7500 | 8200 | 8600 | 9200 | 9700 | 10400
temp | 61 | 58 | 56 | 53 | 50 | 46
a. what is the predicted temperature for an altitude of 6800ft? ______

Explanation:

Step1: Find the slope

First, we take two points from the table. Let's use (7500, 61) and (8200, 58). The slope \( m \) is calculated as \( m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{58 - 61}{8200 - 7500}=\frac{- 3}{700}\approx - 0.00429\) (we can also check other points, like (8200,58) and (8600,56): \(m=\frac{56 - 58}{8600 - 8200}=\frac{-2}{400}=-0.005\), (8600,56) and (9200,53): \(m=\frac{53 - 56}{9200 - 8600}=\frac{-3}{600}=-0.005\), (9200,53) and (9700,50): \(m=\frac{50 - 53}{9700 - 9200}=\frac{-3}{500}=-0.006\), (9700,50) and (10400,46): \(m=\frac{46 - 50}{10400 - 9700}=\frac{-4}{700}\approx - 0.00571\). Maybe a better way is to see the difference in altitude and temperature. From 7500 to 8200 (700 ft increase), temp drops 3. From 8200 to 8600 (400 ft increase), temp drops 2. From 8600 to 9200 (600 ft increase), temp drops 3. From 9200 to 9700 (500 ft increase), temp drops 3. From 9700 to 10400 (700 ft increase), temp drops 4. Alternatively, let's assume a linear relationship (since it's a prediction, linear regression or linear approximation). Let's take two points with a more consistent slope. Let's use (7500, 61) and (9200, 53). The change in altitude \( \Delta x=9200 - 7500 = 1700\), change in temperature \( \Delta y=53 - 61=-8\). So slope \( m=\frac{-8}{1700}\approx - 0.0047\). But maybe a better approach is to see the difference between 7500 and 6800. The altitude decreases by \(7500 - 6800 = 700\) ft. From the first two points, when altitude increases by 700 (from 7500 to 8200), temperature decreases by 3. So when altitude decreases by 700 (from 7500 to 6800), temperature should increase by 3. So starting from 61 (at 7500), increasing by 3 gives \(61 + 3=64\). Wait, let's check the differences:

Altitude: 7500, 8200 (diff 700), 8600 (diff 400 from 8200), 9200 (diff 600 from 8600), 9700 (diff 500 from 9200), 10400 (diff 700 from 9700)

Temperature: 61, 58 (diff -3), 56 (diff -2), 53 (diff -3), 50 (diff -3), 46 (diff -4)

The first interval (7500 to 8200) is 700 ft, temp drops 3. So the rate is 3°F per 700 ft increase. So for a decrease of 700 ft (from 7500 to 6800), temp should increase by 3°F. So 61 + 3 = 64. Let's verify with another interval. From 8200 (8200) to 7500 (7500) is -700 ft, temp goes from 58 to 61, which is +3. So yes, the pattern here is that a 700 ft decrease in altitude leads to a 3°F increase in temperature. So 7500 - 6800 = 700 ft decrease. So temperature at 6800 is 61 + 3 = 64.

Step2: Confirm the prediction

We can also set up a linear equation. Let \( y = mx + b \), where \( y \) is temperature, \( x \) is altitude. Using (7500, 61) and (8200, 58). \( m=\frac{58 - 61}{8200 - 7500}=\frac{-3}{700}\). Then \( 61=\frac{-3}{700}(7500)+b \). Calculate \( \frac{-3\times7500}{700}=\frac{-22500}{700}\approx - 32.14\). So \( b = 61 + 32.14\approx93.14\). Now, for \( x = 6800 \), \( y=\frac{-3}{700}(6800)+93.14=\frac{-20400}{700}+93.14\approx - 29.14 + 93.14 = 64\). Which matches our previous calculation.

Answer:

64