To solve for the final score:

Add the five past test scores and you get 454. Then set up an algebraic equation where you add 454 to $\displaystyle x$, which is the final test score, and divide by six, because you want the average for 6 tests now. You make this equation equal to 90 because that is the average Michael wants and solve for $\displaystyle x$:

$\displaystyle \left ( 454+x\right )\div6 = 90$

$\displaystyle 454+x = 540$

$\displaystyle x=86$

$\displaystyle Distance=rate\times time$ 

Plug in the the values for distance and time, and solve for rate.

$\displaystyle 450 = r\times 6$ 

and 

$\displaystyle r = 75$

The three numbers would be

$\displaystyle x$, $\displaystyle x+2$, and $\displaystyle x+4$.

Add the first and third value and you get

$\displaystyle 2x+4 = 28$ 

and

$\displaystyle x=12$.

The second largest value is $\displaystyle x+2=14$.

Using $\displaystyle Distance = rate\times time$, the time would be $\displaystyle 8.333$ hours, which is $\displaystyle 8$ hours and $\displaystyle 20$ minutes. If you add that to 1pm, you get 9:20pm.

You first find the rate at which the bottle is being filled at, which is  

$\displaystyle 1 oz. -0.2 oz = 0.8 oz$.

Then you divide the entire bottle, which is $\displaystyle 16 oz.$ by the rate of $\displaystyle .8 oz$, and you get $\displaystyle 20\ seconds$. 

To get the final price, you use $\displaystyle x$ as the variable for the original price. So,

$\displaystyle 0.8x = \$ $88$, and divide by 0.8 on both sides of the equations, and you get

$\displaystyle \$110$ for the original price of the shirt.

Look at the work rates as "tanks per minute", not "minutes per tank".

The two pipes can fill the tank up at $\displaystyle \frac{1}{45}$ tanks per minute and  $\displaystyle \frac{1}{40}$ tanks per minute. 

Let $\displaystyle t$ be the time it took, in minutes, to fill the tank up. Then, since rate multiplied by time is equal to work, then two pipes filled up $\displaystyle \frac{1}{45} t$ and $\displaystyle \frac{1}{40} t$ tanks; together, they filled up $\displaystyle \frac{1}{45} t + \frac{1}{40} t$ tank - one tank. This sets up the equation to be solved:

$\displaystyle \frac{1}{45} t + \frac{1}{40} t = 1$

$\displaystyle 360 \cdot \left ( \frac{1}{45} t + \frac{1}{40} t \right ) = 360 \cdot 1$

$\displaystyle \frac{360}{45} t + \frac{360}{40} t = 360$

$\displaystyle 8 t +9 t = 360$

$\displaystyle 17 t = 360$

$\displaystyle t = 360 \div 17 \approx 21.2$ minutes

Look at the work rates as "tanks per minute", not "minutes per tank".

The inlet pipe can fill the tank at a rate of $\displaystyle \frac{1}{25}$ tank per minute. The drain is emptying the tank at a rate of $\displaystyle \frac{1}{55}$ tank per minute.

Let $\displaystyle t$ be the time it took, in minutes, to fill the tank up. Then, since rate multiplied by time is equal to work, then the inlet pipe let in $\displaystyle \frac{1}{25} t$ tank of water, but the drain let out $\displaystyle \frac{1}{55} t$ tank. Since they were working against each other, the work of the latter is subtracted from that of the former, and we can set up and solve the equation:

$\displaystyle \frac{1}{25} t -\frac{1}{55} t = 1$

$\displaystyle 275 \cdot \left ( \frac{1}{25} t -\frac{1}{55} t \right ) = 275 \cdot 1$

$\displaystyle \frac{275 }{25} t -\frac{275 }{55} t = 275$

$\displaystyle 11 t -5 t = 275$

$\displaystyle 6t = 275$

$\displaystyle t = 275 \div 6 \approx 45.8$ minutes

The total of the measures of the angles of a pentagon is

$\displaystyle 180 \cdot (5-2) = 540^{\circ }$.

One angle measures $\displaystyle t ^{\circ }$; since the other four angles have the same measure as one another, we let $\displaystyle X^{\circ }$ be their common measure. Then we set up the equation and solve for $\displaystyle X$:

$\displaystyle 4X + t = 540$

$\displaystyle 4X + t - t = 540- t$

$\displaystyle 4X= 540- t$

$\displaystyle 4X\div 4 = \left ( 540- t \right )\div 4$

$\displaystyle X =135 - \frac{1}{4} t$

Let Julie's score be $\displaystyle x$. Then John's score was $\displaystyle x + 16$. Since the sum of their scores was 80,

$\displaystyle x + x + 16 = 80$

$\displaystyle 2x+16 = 80$

$\displaystyle 2x+16- 16 = 80 - 16$

$\displaystyle 2x = 64$

$\displaystyle 2x \div 2 = 64 \div 2$

$\displaystyle x = 32$

Julie scored 32, and John scored 16 higher, or 48.

Let David's score be $\displaystyle y$. Then Dana's score was $\displaystyle y + 10$. Since the sum of their scores was 60,

$\displaystyle y + y + 10 = 60$

$\displaystyle 2y + 10 = 60$

$\displaystyle 2y + 10 - 10 = 60 - 10$

$\displaystyle 2y = 50$

$\displaystyle 2y\div 2 = 50 \div 2$

$\displaystyle y = 25$

David's score was 25, and Dana's score was 10 higher, or 35.

In descending order of score, the students were John, Dana, Julie, David.