Notice that the number of remaining packages decreases by $30$ every $2$ hours. In other words, the number of remaining packages decreases by \[ \dfrac{30}{2}=15 \] packages per hour. So, if $A$ was the number of packages that Mark started with, then we can model the number of remaining packages $P$ after $n$ hours using the equation

\[ P = A - 15 n. \]

Now, we substitute one of the pairs of values from the table into this equation and solve for $A.$ After $n=4$ hours, there were $P=64$ packages remaining. We substitute these values into the equation and solve for $A,$ as follows:

\begin{align} P &= A - 15 n \\ 64 &= A - 15 (4) \\ 64 &= A - 60 \\ 124 &= A \end{align}

So, Mark started with $124$ packages.

Notice that the loan decreases by $\$ 1\,000$ every $5$ months. In other words, the loan decreases by \[ \dfrac{1\,000}{5}=200 \] dollars per month. So, if $A$ was the amount of money that Ross borrowed, then we can model the amount $L$ outstanding on the loan after $n$ months using the equation

\[ L = A - 200 n. \]

Now, we substitute one of the pairs of values from the table into this equation and solve for $A.$ After $n=10$ months, there was $L = 4\,000$ dollars remaining on the loan. We substitute these values into the equation and solve for $A,$ as follows:

\begin{align} L &= A - 200 n \\ 4\,000 &= A - 200 (10)\\ 4\,000 &= A - 2\,000\\ 6\,000 &= A \end{align}

So, Ross borrowed $\$ 6\,000.$

The model is an equation of a form \[ \text{(starting value)} + \text{(rate of change)}n. \]

We see that the number $3\,000$ corresponds to the starting value, which represents the amount of money that Anna initially borrowed from the bank.

The model is an equation of a form \[ \text{(starting value)} + \text{(rate of change)}m. \]

We see that the number $1.5$ corresponds to the rate of change, which represents the length that hair grows each month.

The model is an equation of a form \[ \text{(starting value)} + \text{(rate of change)}n. \]

The number $40$ corresponds to the rate of change, representing the volume of water coming out of the tank per hour.

The linear function that models the water usage, $W,$ for $t$ years after $2018,$ is \begin{align*} W &= mt + b . \end{align*} where $m$ is the slope, and $b$ is the $W$-intercept.

We have two known points: \begin{align*} \text{for $2018$:} \qquad (t_1, W_1) &= (0, 2400) \\[5pt] \text{for $2023$:} \qquad (t_2, W_2) &= (5, 4200) \end{align*}

We first find the slope $m{:}$ \begin{align*} m &= \frac{W_2 - W_1}{t_2 - t_1} \\[5pt] &= \frac{4,200 - 2,400}{5 - 0} \\[5pt] &= \frac{1,800}{5} \\[5pt] &= 360. \end{align*}

Then, since the water usage in $2018$ (when $t = 0$) is $2,400,$ we have $b = 2,400.$ So we get \begin{align*} W &= mt + b \\[5pt] &= 360t + 2,400. \end{align*}

The linear function that models the population, $P,$ for $t$ years after $2005,$ is

\begin{align*} P &= mt + b . \end{align*}

where $m$ is the slope, and $b$ is the $P$-intercept.

We have two known points:

\begin{align*} \text{for $2005$:} \qquad (t_1, P_1) &= (0, 12\,000) \\[5pt] \text{for $2020$:} \qquad (t_2, P_2) &= (15, 8400) \end{align*}

We first find the slope $m{:}$

\begin{align*} m &= \frac{P_2 - P_1}{t_2- t_1} \\[5pt] &= \frac{8,400 - 12,000}{15 - 0} \\[5pt] &= \frac{-3,600}{15} \\[5pt] &= -240 \end{align*}

Then, since the bird's population in $2005$ (when $t=0$) is $12,000,$ we have $b = 12,000.$ So we get

\begin{align*} P &= mt + b \\[5pt] &= -240t + 12,000. \end{align*}

We can model the number of trailer views $V$ as a linear function of the number of ads $a$ using the equation

\[ V = ma + b, \] where $m$ is the slope and $b$ is the $V$-intercept.

We are given two data points:

\begin{align*} (a_1, V_1) &= (4, 16\,000) \\[5pt] (a_2, V_2) &= (10, 28\,000) \end{align*}

We begin by calculating the slope $m{:}$

\begin{align*} m &= \frac{V_2 - V_1}{a_2 - a_1} \\[5pt] &= \frac{28,000 - 16,000}{10 - 4} \\[5pt] &= \frac{12,000}{6} \\[5pt] &= 2,000 \end{align*}

Now we substitute this value of $m$ into the general form:

\begin{align*} V &= ma + b \\[5pt] &= 2,000a + b \end{align*}

Next, we use the point $(4, 16\,000)$ to solve for $b{:}$

\begin{align*} 16,000 &= 2,000 \times 4 + b \\[5pt] 16,000 &= 8,000 + b \\[5pt] b &= 8,000 \end{align*}

So the linear model is

\[ V = 2,000a + 8,000. \]

To find the number of views when $a = 7{:}$

\begin{align*} V &= 2,000 \times 7 + 8,000 \\[5pt] &= 14,000 + 8,000 \\[5pt] &= 22,000. \end{align*}

The linear equation that models the number of bookings $B,$ in bookings, as a function of tour price $p,$ in dollars, is

\[ B = mp + b, \] where $m$ is the slope, and $b$ is the $B$-intercept.

We have two known data points:

\begin{align*} (p_1, B_1) = (800, 480) \\[5pt] (p_2, B_2) = (1{\,}000, 360) \end{align*}

We first find the slope $m{:}$

\begin{align*} m &= \frac{B_2 - B_1}{p_2 - p_1} \\[5pt] &= \frac{360 - 480}{1{,}000 - 800} \\[5pt] &= \frac{-120}{200} \\[5pt] &= -0.6 \end{align*}

So, the bookings function is

\begin{align*} B &= mp + b \\[5pt] &= -0.6p + b. \end{align*}

Now, we use one of the given points, say $(800, 480),$ to find the value of $b{:}$

\begin{align*} 480 &= -0.6\times 800 + b \\[5pt] 480 &= -480 + b \\[5pt] b &= 960 \end{align*}

So we have

\[ B = -0.6p + 960. \]

To find the price when $B = 420,$ we substitute and solve for $p{:}$

\begin{align*} 420 &= -0.6p + 960 \\[5pt] 420 - 960 &= -0.6p + 960 - 960 \\[5pt] -540 &= -0.6p \\[5pt] 0.6p &= 540 \\[5pt] p &= \frac{540}{0.6} \\[5pt] p &= 900. \end{align*}

Therefore, the tour price would be $\$900.$