If order matters, then there are 12 ways to do this
If order does not matter, then there are 6 ways to do this
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We have 4 choices for the first slot and 3 choices for the next (we can't reuse a letter) so that's where 4*3 = 12 comes from
If order doesn't matter, then something like AB is the same as BA. So we are doubly counting each possible combo. To fix this, we divide by 2: 12/2 = 6
To be more formal, you can use nPr and nCr to get 12 and 6 respectively (use n = 4 and r = 2)
The absolute change in price was $12−$8=$4. Relative to the original price, it was 412, which is 0.3333 or 33.33%.
The answer was 1,150.0 but as a whole number its 1,150.
Answer:
If we mean - √9, then it is a rational number.
Step-by-step explanation:
<span>13-((4/5)+(6/8))
Make your fractions have common denominators
</span>13-((32/40)+(30/40))
Add your fractions and simplify
13-(62/40)
or
13-(31/20)
or
13-(1 11/20)
Then turn 13 into a fraction with a common denominator! Im going to use the second fraction method (31/20)
13 written as a fraction is 13/1, its LCD with 31/20 is 20. I now multiply the top and bottom by 20
260/20
Now I rewrite the problem again
(260/20)-(31/20)
Which equals
229/20!
This is your unsimplified answer
Finally you simplify and get
11 9/20
Answer:
<em>The fraction of the beads that are red is</em>
Step-by-step explanation:
<u>Algebraic Expressions</u>
A bag contains red (r), yellow (y), and blue (b) beads. We are given the following ratios:
r:y = 2:3
y:b = 5:4
We are required to find r:s, where s is the total of beads in the bag, or
s = r + y + b
Thus, we need to calculate:
![\displaystyle \frac{r}{r+y+b} \qquad\qquad [1]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Br%7D%7Br%2By%2Bb%7D%20%20%20%20%20%20%20%5Cqquad%5Cqquad%20%20%20%20%5B1%5D)
Knowing that:
![\displaystyle \frac{r}{y}=\frac{2}{3} \qquad\qquad [2]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Br%7D%7By%7D%3D%5Cfrac%7B2%7D%7B3%7D%20%20%20%20%20%20%5Cqquad%5Cqquad%20%20%20%20%5B2%5D)
Multiplying the equations above:
Simplifying:
![\displaystyle \frac{r}{b}=\frac{5}{6} \qquad\qquad [3]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Br%7D%7Bb%7D%3D%5Cfrac%7B5%7D%7B6%7D%20%20%20%20%20%20%20%5Cqquad%5Cqquad%20%20%20%20%5B3%5D)
Dividing [1] by r:
Substituting from [2] and [3]:
Operating:
The fraction of the beads that are red is 
