Answer:
![1 \to 22 \to 0.176](https://tex.z-dn.net/?f=1%20%5Cto%2022%20%5Cto%200.176)
![2 \to 13 \to 0.104](https://tex.z-dn.net/?f=2%20%5Cto%2013%20%5Cto%200.104)
![3 \to 18 \to 0.144](https://tex.z-dn.net/?f=3%20%5Cto%2018%20%5Cto%200.144)
![4 \to 29 \to 0.232](https://tex.z-dn.net/?f=4%20%5Cto%2029%20%5Cto%200.232)
![5 \to 37 \to 0.296](https://tex.z-dn.net/?f=5%20%5Cto%2037%20%5Cto%200.296)
![6 \to 6 \to 0.048](https://tex.z-dn.net/?f=6%20%5Cto%206%20%5Cto%200.048)
Step-by-step explanation:
Given
![n = 125](https://tex.z-dn.net/?f=n%20%3D%20125)
See attachment for proper table
Required
Complete the table
Experimental probability is calculated as:
![Pr = \frac{Frequency}{n}](https://tex.z-dn.net/?f=Pr%20%3D%20%5Cfrac%7BFrequency%7D%7Bn%7D)
We use the above formula when the frequency is known.
For result of roll 2, 4 and 6
The frequencies are 13, 29 and 6, respectively
So, we have:
![Pr(2) = \frac{13}{125} = 0.104](https://tex.z-dn.net/?f=Pr%282%29%20%3D%20%5Cfrac%7B13%7D%7B125%7D%20%3D%200.104)
![Pr(4) = \frac{29}{125} = 0.232](https://tex.z-dn.net/?f=Pr%284%29%20%3D%20%5Cfrac%7B29%7D%7B125%7D%20%3D%200.232)
![Pr(6) = \frac{6}{125} = 0.048](https://tex.z-dn.net/?f=Pr%286%29%20%3D%20%5Cfrac%7B6%7D%7B125%7D%20%3D%200.048)
When the frequency is to be calculated, we use:
![Pr = \frac{Frequency}{n}](https://tex.z-dn.net/?f=Pr%20%3D%20%5Cfrac%7BFrequency%7D%7Bn%7D)
![Frequency = n * Pr](https://tex.z-dn.net/?f=Frequency%20%3D%20n%20%2A%20Pr)
For result of roll 3 and 5
The probabilities are 0.144 and 0.296, respectively
So, we have:
![Frequency(3) = 125 * 0.144 = 18](https://tex.z-dn.net/?f=Frequency%283%29%20%3D%20125%20%2A%200.144%20%3D%2018)
![Frequency(5) = 125 * 0.296 = 37](https://tex.z-dn.net/?f=Frequency%285%29%20%3D%20125%20%2A%200.296%20%3D%2037)
For roll of 1 where the frequency and the probability are not known, we use:
![Total \ Frequency = 125](https://tex.z-dn.net/?f=Total%20%5C%20Frequency%20%3D%20125)
So:
Frequency(1) added to others must equal 125
This gives:
![Frequency(1) + 13 + 18 + 29 + 37 + 6 = 125](https://tex.z-dn.net/?f=Frequency%281%29%20%2B%2013%20%2B%2018%20%2B%2029%20%2B%2037%20%2B%206%20%3D%20125)
![Frequency(1) + 103 = 125](https://tex.z-dn.net/?f=Frequency%281%29%20%2B%20103%20%3D%20125)
Collect like terms
![Frequency(1) =- 103 + 125](https://tex.z-dn.net/?f=Frequency%281%29%20%3D-%20103%20%2B%20125)
![Frequency(1) =22](https://tex.z-dn.net/?f=Frequency%281%29%20%3D22)
The probability is then calculated as:
![Pr(1) = \frac{22}{125}](https://tex.z-dn.net/?f=Pr%281%29%20%3D%20%5Cfrac%7B22%7D%7B125%7D)
![Pr(1) = 0.176](https://tex.z-dn.net/?f=Pr%281%29%20%3D%200.176)
So, the complete table is:
![1 \to 22 \to 0.176](https://tex.z-dn.net/?f=1%20%5Cto%2022%20%5Cto%200.176)
![2 \to 13 \to 0.104](https://tex.z-dn.net/?f=2%20%5Cto%2013%20%5Cto%200.104)
![3 \to 18 \to 0.144](https://tex.z-dn.net/?f=3%20%5Cto%2018%20%5Cto%200.144)
![4 \to 29 \to 0.232](https://tex.z-dn.net/?f=4%20%5Cto%2029%20%5Cto%200.232)
![5 \to 37 \to 0.296](https://tex.z-dn.net/?f=5%20%5Cto%2037%20%5Cto%200.296)
![6 \to 6 \to 0.048](https://tex.z-dn.net/?f=6%20%5Cto%206%20%5Cto%200.048)
Answer:
535.9 ft²
Step-by-step explanation:
Since there are 360° in a circle, and 240° is 2/3 of 360°, we can say that the area of the bolded sector is <em>2/3 the area of the whole circle</em>. The area of a circle with a radius of r is πr², or approximately 3.14r², so the area of the whole circle is ≈ 3.14(16)² = 3.14(256) = 803.84 ft². Taking 2/3 of this gets us 803.84 * (2/3) ≈ 535.9 ft²
(X+12) + 140 = 180
X + 12 + 140 = 180
X + 152 = 180
X = 180 - 152
X = 28
You have to make 4% into a decimal which is.04 then multiply 15600 x .04 which equals 624