Like I said above, There were 40 rolls, 6 of which were a two (6/40) and 8 of which were a 4 (8/40). Simply add these together to get 14/40 as the probability of rolling a 2 or 4. This simplifies to 7/20
A continuous data set is where you can plot every point (even very, very small decimal numbers) because the unit can be measured in parts of the whole.
A discrete data set is where you can only plot the units in whole values. For example, if you were plotting the number of people, you can't have a part of a person.
Weight is measured in lbs which can be measured in parts of the lb (such as 100.005 lbs) so it is continuous.
Answer: Continuous data set
ANSWER: 32 five-dollar bills
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EXPLANATION:
Let x be number of $5 bills
Let y be number of $10 bills
Since we have total of 38 bills, we must have the sum of x and y be 38
x + y = 38 (I)
Since the total amount deposited is $220, we must have the sum of 5x and 10y be 220 (x and y are just the "number of" their respective bills, so we multiply them by their value to get the total value):
5x + 10y = 220 (II)
System of equations:
![\left\{ \begin{aligned} x + y &= 38 && \text{(I)} \\ 5x + 10y &= 220 && \text{(II)} \end{aligned} \right.](https://tex.z-dn.net/?f=%20%5Cleft%5C%7B%20%5Cbegin%7Baligned%7D%20x%20%2B%20y%20%26%3D%2038%20%26%26%20%5Ctext%7B%28I%29%7D%20%5C%5C%205x%20%2B%2010y%20%26%3D%20220%20%26%26%20%5Ctext%7B%28II%29%7D%20%5Cend%7Baligned%7D%20%5Cright.%20)
Divide both sides of equation (II) by 5 so our numbers become smaller
![\left\{ \begin{aligned} x + y &= 38 && \text{(I)} \\ x + 2y &= 44 && \text{(II)} \end{aligned} \right.](https://tex.z-dn.net/?f=%20%5Cleft%5C%7B%20%5Cbegin%7Baligned%7D%20x%20%2B%20y%20%26%3D%2038%20%26%26%20%5Ctext%7B%28I%29%7D%20%5C%5C%20x%20%2B%202y%20%26%3D%2044%20%26%26%20%5Ctext%7B%28II%29%7D%20%5Cend%7Baligned%7D%20%5Cright.%20)
Rearrange (I) to solve for y so that we can substitute into (II)
![\begin{aligned} x + y &= 38 && \text{(I)} \\ y &= 38 - x \end{aligned}](https://tex.z-dn.net/?f=%20%5Cbegin%7Baligned%7D%20x%20%2B%20y%20%26%3D%2038%20%26%26%20%5Ctext%7B%28I%29%7D%20%5C%5C%20y%20%26%3D%2038%20-%20x%20%5Cend%7Baligned%7D%20)
Substituting this into equation (II) for the y:
![\begin{aligned} x + 2y &= 44 && \text{(II)} \\ x + 2(38 - x) &= 44\\ x + 76 - 2x &= 44 \\ -x &= -32 \\ x &= 32 \end{aligned}](https://tex.z-dn.net/?f=%20%5Cbegin%7Baligned%7D%20x%20%2B%202y%20%26%3D%2044%20%26%26%20%5Ctext%7B%28II%29%7D%20%5C%5C%20x%20%2B%202%2838%20-%20x%29%20%26%3D%2044%5C%5C%20x%20%2B%2076%20-%202x%20%26%3D%2044%20%5C%5C%20-x%20%26%3D%20-32%20%5C%5C%20x%20%26%3D%2032%20%5Cend%7Baligned%7D%20)
We have 32 five-dollar bills
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If we want to finish off the question, use y = 38 - x to figure out number of $10 bills
![y = 38 - 32 = 6](https://tex.z-dn.net/?f=y%20%3D%2038%20-%2032%20%3D%206)
32 five-dollar bills and 6 ten-dollar bills
I know that x= 60. I don't know what a proportion is. sorry.
Perimeter (P) = 2L + 2w
68 = 2L + 2w
68 - 2L = 2w
2(34 - L) = 2(w)
34 - L = w
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Area (A) = L * w
240 = L (34 - L)
240 = 34L - L²
L² - 34L + 240 = 0
(L - 10)(L - 24) = 0
L = 10 or L = 24
w = 34 - L = 34 - 10 = 24 or w = 34 - 24 = 10
Answer: width = 10, length = 24 assuming length is bigger than the width