Question

when the air temperature reaches the dew - point, fog may form. this phenomenon also causes clouds to form at higher altitudes. both the air temperature and the dew - point decrease at a (truncated) as the altitude above ground level increases. if the ground - level temperature and dew - point are (t_0) and (d_0) respectively, the air temperature at an altitude of (x) miles can be approximated by (t(x)=t_0 - 19x), and the dew - point can be approximated by (d(x)=d_0 - 5.8x). suppose the ground - level temperature is (62^{circ}f) and the dew - point is (47^{circ}f). note that clouds will not form at altitudes for which the air temperature is above the dew - point. (a) use the intersection - of - graphs method to estimate the altitudes at which clouds will not form. (b) solve part (a) analytically. (a) the altitudes at which clouds will not form is below 1.12 mi (round to the nearest hundredth as needed.) (b) what is the solution set, in interval notation, when the problem is solved analytically? a. ((-infty,infty)) b. (0,square) (type an integer or a simplified fraction.) c. (varnothing) d. (square,infty)) (type an integer or a simplified fraction.)

Explanation:

Step1: Set up the equations

Let $T_0 = 62^{\circ}F$ be the ground - level temperature and $D_0=47^{\circ}F$ be the ground - level dew - point. The air temperature function is $T(x)=T_0 - 19x$ and the dew - point function is $D(x)=D_0 - 5.8x$, where $x$ is the altitude in miles. Clouds do not form when $T(x)>D(x)$. So, we set up the inequality $T_0 - 19x>D_0 - 5.8x$.

Step2: Substitute the values of $T_0$ and $D_0$

Substitute $T_0 = 62$ and $D_0 = 47$ into the inequality: $62-19x>47 - 5.8x$.

Step3: Solve the inequality for $x$

First, add $19x$ to both sides: $62>47 - 5.8x+19x$. Then simplify to get $62>47 + 13.2x$. Next, subtract 47 from both sides: $62 - 47>13.2x$, which gives $15>13.2x$. Finally, divide both sides by 13.2: $x<\frac{15}{13.2}\approx1.14$ miles. In interval notation, the solution set is $(-\infty, \frac{15}{13.2})=(-\infty,1.14)$. Since we want the set where clouds do not form, and we know from part (a) that clouds do not form below a certain altitude, the solution set in interval notation is $(-\infty, \frac{15}{13.2})\approx(-\infty,1.14)$. The correct answer for part (b) is the set of all real numbers less than the value of $x$ where $T(x) = D(x)$. Solving $T(x)=D(x)$ gives:
\[

$$\begin{align*} T_0 - 19x&=D_0 - 5.8x\\ 62-19x&=47 - 5.8x\\ -19x + 5.8x&=47 - 62\\ -13.2x&=-15\\ x&=\frac{15}{13.2}\approx1.14 \end{align*}$$

\]
The solution set in interval notation is $(-\infty,\frac{15}{13.2})$. Rounding $\frac{15}{13.2}\approx1.14$. The solution set in interval notation for the values of $x$ where $T(x)>D(x)$ is $(-\infty, \frac{15}{13.2})$. In the given options, if we consider the non - rounded value, the answer is $(-\infty,\frac{15}{13.2})$. If we assume the options are asking for a more "clean" form, we note that when we solve $62-19x>47 - 5.8x$ we get $x < \frac{15}{13.2}\approx1.14$. The solution set in interval notation is $(-\infty,1.14)$. If we want an integer or fraction form for the upper - bound, $x<\frac{15}{13.2}=\frac{150}{132}=\frac{25}{22}\approx1.14$.

Answer:

D. $(-\infty,\frac{25}{22})$