Questions: The temperature at which water starts to boil is called its boiling point and is linearly related to the altitude. Water boils at 212 degrees F at sea level and at 193.6 degrees F at an altitude of 10,000 feet. (a) Find a relationship of the form T=mx+b where T is degrees Fahrenheit and x is altitude in thousands of feet. (b) Find the boiling point at an altitude of 3,800 feet. (c) Find the altitude if the boiling point is 204.5 degrees F. (d) Graph T and illustrate the answers to (b) and (c) on the graph. (a) Choose the correct relationship below. A. T=1.84 x+212 B. T=-1.84 x-212 C. T=1.84 x-212 D. T=-1.84 x+212 (b) The boiling point at an altitude of 3,800 feet is degrees F. (Round to two decimal places as needed.)

The temperature at which water starts to boil is called its boiling point and is linearly related to the altitude. Water boils at 212 degrees F at sea level and at 193.6 degrees F at an altitude of 10,000 feet. (a) Find a relationship of the form T=mx+b where T is degrees Fahrenheit and x is altitude in thousands of feet. (b) Find the boiling point at an altitude of 3,800 feet. (c) Find the altitude if the boiling point is 204.5 degrees F. (d) Graph T and illustrate the answers to (b) and (c) on the graph. (a) Choose the correct relationship below. A. T=1.84 x+212 B. T=-1.84 x-212 C. T=1.84 x-212 D. T=-1.84 x+212 (b) The boiling point at an altitude of 3,800 feet is degrees F. (Round to two decimal places as needed.)
Transcript text: The temperature at which water starts to boil is called its boiling point and is linearly related to the altitude. Water boils at $212^{\circ} \mathrm{F}$ at sea level and at $193.6^{\circ} \mathrm{F}$ at an altitude of 10,000 feet. (a) Find a relationship of the form $\mathrm{T}=\mathrm{mx}+\mathrm{b}$ where T is degrees Fahrenheit and x is altitude in thousands of feet. (b) Find the boiling point at an altitude of 3,800 feet. (c) Find the altitude if the boiling point is $204.5^{\circ} \mathrm{F}$. (d) Graph $T$ and illustrate the answers to (b) and (c) on the graph. (a) Choose the correct relationship below. A. $T=1.84 x+212$ B. $T=-1.84 x-212$ C. $T=1.84 x-212$ D. $T=-1.84 x+212$ (b) The boiling point at an altitude of 3,800 feet is $\square$ ${ }^{\circ} \mathrm{F}$. (Round to two decimal places as needed.)
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Solution

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Solution Steps

Step 1: Calculate the slope (m)

Given two points: (x1, T1) = (0, 212) and (x2, T2) = (10, 193.6), the slope \(m\) can be calculated as \(m = (T_2 - T_1) / (x_2 - x_1) = (193.6 - 212) / (10 - 0) = -1.84\).

Step 2: Calculate the y-intercept (b)

Using the point (x1, T1) = (0, 212) and the slope \(m\), the y-intercept \(b\) can be calculated as \(b = T1 - m \times x1 = 212 - (-1.84) \times 0 = 212\).

Step 3: Solve for boiling point (T)

Given the altitude \(x = 3.8\), solve for \(T\) using \(T = mx + b\). Substituting the known values, \(T = -1.84 \times 3.8 + 212 = 205.01\).

Final Answer: The boiling point of the liquid at 3.8 feet altitude is approximately 205.01°.

Step 1: Calculate the slope (m)

Given two points: (x1, T1) = (0, 212) and (x2, T2) = (10, 193.6), the slope \(m\) can be calculated as \(m = (T_2 - T_1) / (x_2 - x_1) = (193.6 - 212) / (10 - 0) = -1.84\).

Step 2: Calculate the y-intercept (b)

Using the point (x1, T1) = (0, 212) and the slope \(m\), the y-intercept \(b\) can be calculated as \(b = T1 - m \times x1 = 212 - (-1.84) \times 0 = 212\).

Step 3: Solve for altitude (x)

Given the temperature \(T = 204.5\), solve for \(x\) using \(x = (T - b) / m\). Substituting the known values, \(x = (204.5 - 212) / -1.84 = 4.08\).

Final Answer: The altitude at which the liquid boils at 204.5° is approximately 4.08 feet.
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