1/3 of the movies are action movies. 3/5 of 2/3 of the movies are comedies, or (3/5)*(2/3), or 6/15. Combining the comedies and the action movies (1/3 or 5/15), we get 11/15 of the movies being either action or comedy. Thus, 4/15 of the movies remain and all of them have to be historical.

To answer this question, we need to convert $\displaystyle 18\%$ to a decimal. $\displaystyle 18\%=0.18$. Now we multiply $\displaystyle 1,200$ by $\displaystyle 0.18$. $\displaystyle 1200\cdot 0.18=216$.

In order to solve this, we need to set up an equation. $\displaystyle 549\:\text{m}=\frac{10\:\text{m}}{\text{sec}}\cdot x$, where $\displaystyle x$ is time. All we need to do is divide by $\displaystyle 10\: \frac{\text{m}}{\text{sec}}$ on each side.

$\displaystyle 549\:\text{m}=\frac{10\:\text{m}}{\text{s}}\cdot x$

$\displaystyle \frac{549\:\text{m}}{\frac{10\:\text{m}}{\text{sec}}}=x$

$\displaystyle \frac{549\:\text{m\:sec}}{10\:\text{m}}=x$

$\displaystyle x=\frac{549\:\text{sec}}{10}=54.9\:\text{sec}$

First we need to create equations that represent the path's of the Rocket's.

For Rocket $\displaystyle A$, the equation is

$\displaystyle y=25x$

For Rocket $\displaystyle B$, the equation is

$\displaystyle y=16x+3$

In order to solve for the time the Rocket's cross, we need to set the equations equal to each other.

$\displaystyle 25x=16x+3$

Now solve for $\displaystyle x$

$\displaystyle 9x=3$

$\displaystyle x=\frac{1}{3}$

To figure out which month had a higher approval rate, we need to divide the approve number by the total number of people.

$\displaystyle \text{September Approval Rate}= \frac{345}{1500}=0.23=23\%$

$\displaystyle \text{October Approval Rate}= \frac{440}{1500}=0.29=29\%$

From this, we can determine that October had the highest approval rate.

In order to solve for the slope, we need to recall how to find the slope of a line. $\displaystyle \text{Slope}=\frac{y_2-y_1}{x_2-x_1}$, where $\displaystyle (x_1,y_1), (x_2, y_2)$ are points on the line.

So we will pick $\displaystyle (-5, 80)$, and $\displaystyle (5, -60)$.

$\displaystyle \text{Slope}=\frac{-60-80}{5-(-5)}=\frac{-140}{10}=-14$

To determine the mode, we need to count how many diamonds are on each number. The mode is the most frequently occurring number. 

Since $\displaystyle 10$ has a frequency of $\displaystyle 5$, it is the mode.

To calculate the mean, we need to sum up every individual entry and divide it by the total number of entries. This looks like $\displaystyle \bar{x}=\frac{x_1+x_2+\cdots x_n}{n}$, where $\displaystyle n$ is the total number of entries and $\displaystyle x_1,\cdots, x_n$ are the individual entries.

Here is the calculation of the mean,

$\displaystyle \bar{x}=\frac{3(4)+5(2)+6(2)+7(1)+8(3)+9(3)+10(5)}{20}$

$\displaystyle \bar{x}=\frac{12+10+12+7+24+27+50}{20}$

$\displaystyle \bar{x}=\frac{142}{20}=7.1$

Since we are given the functions x-intercepts, we can write the equation as $\displaystyle f(x)=(x+2)(x-2)(x-1)$, now we need to use FOIL.

Start with $\displaystyle (x+2)(x-2)$.

$\displaystyle (x+2)(x-2)=x^2-2x+2x-4=x^2-4$

Now we do

$\displaystyle (x^2-4)(x-1)=x^3-x^2-4x+4$

The first step is to find the common ratio amongst the data. 

$\displaystyle \frac{\text{Time 1}}{\text{Time 0}}=\frac{500}{1000}=\frac{1}{2}$

$\displaystyle \frac{\text{Time 2}}{\text{Time 1}}=\frac{250}{500}=\frac{1}{2}$

$\displaystyle \frac{\text{Time 3}}{\text{Time 2}}=\frac{125}{250}=\frac{1}{2}$

$\displaystyle \frac{\text{Time 4}}{\text{Time 3}}=\frac{62.5}{125}=\frac{1}{2}$

$\displaystyle \frac{\text{Time 5}}{\text{Time 4}}=\frac{31.25}{62.5}=\frac{1}{2}$

We can that the common ratio is $\displaystyle \frac{1}{2}$.

Now we need to set up an equation that will give an answer for any time.

We can set up an exponential decay model, the general equation is,

$\displaystyle y=c(1-r)^t$, where $\displaystyle c$ is the starting amount, $\displaystyle r$ is the common ratio and $\displaystyle t$ is time.

After plugging our numbers in, we get

$\displaystyle y=1000\left( 1-\frac{1}{2}\right )^t$

$\displaystyle y=1000\left(\frac{1}{2}\right )^t$