Sprint 17904 and solutions

You can pretend the price is $100 and apply the discounts. The number you have at the end is the price you would pay after the discounts.
The single discount then would be
`100 - (\text(price you would pay))`

It doesn't matter what price you choose to start with.The single discount is found by computing
`\text((the price you started with)) - `
`(\text(the price you would pay after applying the discounts to the starting price)) `
We'll compute the answer by assuming the starting price is $100.
A 35% discount on $100 is $35 so we have a new price of $100-35 or $65.
We then compute the second discount:
`$65*34/100 = $22.10`
`$65-$22.10 = $42.90`
$42.90 is the price after the two discounts so the single discount that would give you a final price of $42.90 if you started at $100 is
`$100 - $42.90 `
The answer is 57.10% Notice the "%" follows the number.

The question is what does Kai have to roll to win? Kai has to roll a 7 to win the game.
Count the number of ways Kai can roll a 7. He can roll a

• 6 & 1
• 5 & 2
• 4 & 3
• 3 & 4
• 2 & 5
• 1 & 6

There are 6 ways to roll a 7. There are 36 ways 2 dice can roll so the probability, desired/total, is 6/36.

Reduce the answer to 1/6.

This is a monsoon problem. Just as the Monsoon problem asks how many ways can you arrange the letters in the word Monsoon this question asks ow many ways can you arrange the letters below?

GGGBBBBBBBBBBHHHHHH

The 'G' stands for a single Golden, the 'H' for a single Harcot and the 'B' for a single Bartlett.

There are a total of 19 trees. If they were distinct, there would be `19!` of arranging them. The number of ways of arranging the 3 Goldens is 3!. The number of ways of arranging the 10 Bartletts is 10!. The number of ways of arranging the 6 Harcots is 6!. You correct the `19!` count by dividing each tree's contribution to the total or:
`(19!)/( 3! * 10! * 6! )`

The answer to 3 significant digits is
`7.76*10^6`

Look at the figure shown. Compute the average height in rectangles.

1. `\text(Average height in rectangles)=(3+1)/2`.
   The average height of the figure is 2 rectangles high.
2. Compute the total number of rectangles in the figure.
   `\text(total rectangles in the figure)=`
   `\text((Number of steps wide))*\text((Average number of steps high)) `
   `\text((total rectangles in the figure))=3*2`
   The total number of rectangles is 6. Count them to be sure.
3. Compute the area of a single step that is 2 cm high by 11 cm wide:
4. ` \text(Area of single step)=2*11 ` or 22 `cm^2`
5. Multiply the total number of steps by the area of a single one
6. `text(Total area)= 6*22` or `132 cm^2`.

As your correction exercise, apply the exact same logic to the original problem. Show each step above except substitute the numbers in the problem.

Show that your answer is `67782`. If it isn't, find the error.

The question asks what is the probability that she identified all the correct seats. To answer the question, you have to answer three sub questions.

In class, we use the shorthand ...
`Probability = \text( desired )/ \text( total )`


Of all the ways the conductor can mark the form, there's exactly one way that correctly identifies the broken seats so the number of ways to get the desired outcome is `1`.

Thetotal number of outcomes is found by computing `( \ _28C_3)`

We use Choose instead of Permute because the order the seats are marked doesn't matter. The answer is found by computing

`Probability = 1 / ( \ 28C_3)`



The answer in scientific notation is `3.05*10^(-4)`

Pro Tips!

1. When you calculate `\ _nC_r` or `\ _nP_r` , the correct answer is always a positive integer. If your calculator gives you a negative or fractional answer, you've made a mistake.
2. Probability answers are always a fraction between 0 and 1 inclusive.That's because the desired number of outcomes can't be bigger than the total number of outcomes.

One approach is to solve a simpler problem and then extend the solution.

Consider the cube shown below.

The cube has 4 cubes per edge. The top face shows the 8 cubes which have two painted faces. The cubes on the top face that are not marked are the interior cubes which are painted on just one face and the corner cubes which have three painted faces.

The marked cubes are all on the cube edges. For this example, there are 4 cubes on an edge including the two corner cubes. Subtracting the two corner cubes from the 4 edge cube gives you the number of cubes on an edge that have exactly two painted faces. In this case, each edge has two painted faces.

You need to count the number of edges and multiply by the number of cubes with two painted faces on each edge.

A cube has 4 edges on the top, 4 edges on the bottom and 4 edges on the side. However, we exclude the 4 edges on the bottom becasue the bottom isn't painted in this problem. The total number of edges then is 4 on top plus 4 on the sides or 8.

For the question asked, you need to figure out how many cubes there are on each edge. There are a total of 343 cubes in the problem so each edge has 7 cubes since `7^3=343`.

Subtract 2 from 7 gives 5 cubes on each edge that has two painted faces.

Multiply 5 cubes per edge by 8 edges with two painted faces and the answer is 40 cubes have two painted faces.