Answer: 1/3
Step-by-step explanation:
Here is the complete question:
machine is set to pump cleanser into a process at the rate of 9 gallons per minute. Upon inspection, it is learned that the machine actually pumps cleanser at a rate described by the uniform distribution over the interval 8.5 to 11.5 gallons per minute. Find the probability that between 9.0 gallons and 10.0 gallons are pumped during a randomly selected minute.
The Probability of the above question will be calculated as:
= (10-9) / (11.5 - 8.5)
= 1/3
The above is the easier way to solve this
Is that graph on top of it, for that equation>
Answer:
answer in explanation
Step-by-step explanation:
it is function because each domain is associated with only one range.
all the element of x are domain and y are range.
Answer: The probability that she gets a ride 3 times in a 5-day work week is <u>0.31</u>.
The probability that she gets a ride at least 2 times in a 5-day workweek is <u>0.97</u>.
Step-by-step explanation:
By binomial distribution formula:
![P[X=r]=^nC_rp^rq^{n-r}](https://tex.z-dn.net/?f=P%5BX%3Dr%5D%3D%5EnC_rp%5Erq%5E%7Bn-r%7D)
, where n is the number of trials , r is the number of success, p is the probability of success and q is the probability of failure.
Given : n=5
p = 0.7
q= 1-0.7=0.3
Now, the probability that she gets a ride 3 times in a 5-day work week :
![P[X=3]=^5C_3(0.7)^3(0.3)^{5-3}\\\\=10(0.7)^3(0.3)^2=0.3087\approx0.31](https://tex.z-dn.net/?f=P%5BX%3D3%5D%3D%5E5C_3%280.7%29%5E3%280.3%29%5E%7B5-3%7D%5C%5C%5C%5C%3D10%280.7%29%5E3%280.3%29%5E2%3D0.3087%5Capprox0.31)
The probability that she gets a ride at least 2 times in a 5-day workweek :
![P[X\geq2]=1-P[X](https://tex.z-dn.net/?f=P%5BX%5Cgeq2%5D%3D1-P%5BX%3C2%5D%5C%5C%5C%5C%3D1-%28P%5BX%3D1%5D%2BP%5BX%3D0%5D%29%5C%5C%5C%5C%3D1-%28%5E5C_0%280.7%29%5E0%280.3%29%5E%7B5%7D%2B%5E5C_1%280.7%29%5E1%280.3%29%5E%7B5-1%7D%29%5C%5C%5C%5C%3D1-%5B%280.3%29%5E5%2B5%280.7%29%280.3%29%5E4%5D%5C%5C%5C%5C%3D1-0.03078%3D0.96922%5Capprox0.97)
Formula for the length of x
