Answer:
A. (x*2.5)*x
B. (32*2.5)*32=2560
Step-by-step explanation:
Answer:
Angles 1 and 9 are corresponding angles
2 and 3 same side interior angles
8 and 12 are OTHER
6 and 12 are alternate interior
1 and 2 are linear pairs
15 and 11 are alternate interior
8 and 9 are same side interior
2 and 4 are corresponding
Step-by-step explanation:
Use the graph!
Brainliest??Hope it helped!
Using the binomial distribution, the probabilities are given as follows:
- 0.3675 = 36.75% probability that more than 4 weigh more than 20 pounds.
- 0.1673 = 16.73% probability that fewer than 3 weigh more than 20 pounds.
- Since P(X > 7) < 0.05, it would be unusual if more than 7 of them weigh more than 20 pounds.
<h3>What is the binomial distribution formula?</h3>
The formula is:
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
![C_{n,x} = \frac{n!}{x!(n-x)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
The parameters are:
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
The values of the parameters for this problem are:
n = 10, p = 0.4.
The probability that more than 4 weigh more than 20 pounds is:
![P(X > 4) = 1 - P(X \leq 4)](https://tex.z-dn.net/?f=P%28X%20%3E%204%29%20%3D%201%20-%20P%28X%20%5Cleq%204%29)
In which:
![P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)](https://tex.z-dn.net/?f=P%28X%20%5Cleq%204%29%20%3D%20P%28X%20%3D%200%29%20%2B%20P%28X%20%3D%201%29%20%2B%20P%28X%20%3D%202%29%20%2B%20P%28X%20%3D%203%29%20%2B%20P%28X%20%3D%204%29)
Then:
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
![P(X = 0) = C_{10,0}.(0.4)^{0}.(0.6)^{10} = 0.0061](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20C_%7B10%2C0%7D.%280.4%29%5E%7B0%7D.%280.6%29%5E%7B10%7D%20%3D%200.0061)
![P(X = 1) = C_{10,1}.(0.4)^{1}.(0.6)^{9} = 0.0403](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%20C_%7B10%2C1%7D.%280.4%29%5E%7B1%7D.%280.6%29%5E%7B9%7D%20%3D%200.0403)
![P(X = 2) = C_{10,2}.(0.4)^{2}.(0.6)^{8} = 0.1209](https://tex.z-dn.net/?f=P%28X%20%3D%202%29%20%3D%20C_%7B10%2C2%7D.%280.4%29%5E%7B2%7D.%280.6%29%5E%7B8%7D%20%3D%200.1209)
![P(X = 3) = C_{10,3}.(0.4)^{3}.(0.6)^{7} = 0.2150](https://tex.z-dn.net/?f=P%28X%20%3D%203%29%20%3D%20C_%7B10%2C3%7D.%280.4%29%5E%7B3%7D.%280.6%29%5E%7B7%7D%20%3D%200.2150)
![P(X = 4) = C_{10,4}.(0.4)^{4}.(0.6)^{6} = 0.2502](https://tex.z-dn.net/?f=P%28X%20%3D%204%29%20%3D%20C_%7B10%2C4%7D.%280.4%29%5E%7B4%7D.%280.6%29%5E%7B6%7D%20%3D%200.2502)
Hence:
![P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0061 + 0.0403 + 0.1209 + 0.2150 + 0.2502 = 0.6325](https://tex.z-dn.net/?f=P%28X%20%5Cleq%204%29%20%3D%20P%28X%20%3D%200%29%20%2B%20P%28X%20%3D%201%29%20%2B%20P%28X%20%3D%202%29%20%2B%20P%28X%20%3D%203%29%20%2B%20P%28X%20%3D%204%29%20%3D%200.0061%20%2B%200.0403%20%2B%200.1209%20%2B%200.2150%20%2B%200.2502%20%3D%200.6325)
![P(X > 4) = 1 - P(X \leq 4) = 1 - 0.6325 = 0.3675](https://tex.z-dn.net/?f=P%28X%20%3E%204%29%20%3D%201%20-%20P%28X%20%5Cleq%204%29%20%3D%201%20-%200.6325%20%3D%200.3675)
0.3675 = 36.75% probability that more than 4 weigh more than 20 pounds.
The probability that fewer than 3 weigh more than 20 pounds is:
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0061 + 0.0403 + 0.1209 = 0.1673
0.1673 = 16.73% probability that fewer than 3 weigh more than 20 pounds.
For more than 7, the probability is:
![P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10)](https://tex.z-dn.net/?f=P%28X%20%3E%207%29%20%3D%20P%28X%20%3D%208%29%20%2B%20P%28X%20%3D%209%29%20%2B%20P%28X%20%3D%2010%29)
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
![P(X = 8) = C_{10,8}.(0.4)^{8}.(0.6)^{2} = 0.0106](https://tex.z-dn.net/?f=P%28X%20%3D%208%29%20%3D%20C_%7B10%2C8%7D.%280.4%29%5E%7B8%7D.%280.6%29%5E%7B2%7D%20%3D%200.0106)
![P(X = 9) = C_{10,9}.(0.4)^{9}.(0.6)^{1} = 0.0016](https://tex.z-dn.net/?f=P%28X%20%3D%209%29%20%3D%20C_%7B10%2C9%7D.%280.4%29%5E%7B9%7D.%280.6%29%5E%7B1%7D%20%3D%200.0016)
![P(X = 10) = C_{10,10}.(0.4)^{10}.(0.6)^{0} = 0.0001](https://tex.z-dn.net/?f=P%28X%20%3D%2010%29%20%3D%20C_%7B10%2C10%7D.%280.4%29%5E%7B10%7D.%280.6%29%5E%7B0%7D%20%3D%200.0001)
Since P(X > 7) < 0.05, it would be unusual if more than 7 of them weigh more than 20 pounds.
More can be learned about the binomial distribution at brainly.com/question/24863377
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The number of days d, he must use the gym to make the membership worthwhile is at least 21 days
Equation
Non members:
Cost of swimming per day = $5
Cost of exercise per day = $9
Total = $5 + $9
= $14
Members:
Yearly fee = $300
Exercise fee per day = $4
Swimming fee = $0
The number of days d, he must use the gym to make the membership worthwhile = $300 ÷ 14
= 21.42857142857142
Approximately,
21 days
Learn more about equation:
brainly.com/question/2972832
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Answer:
The answer is 1.178
Step-by-step explanation: