Answer:
33.33% probability that it takes Isabella more than 11 minutes to wait for the bus
Step-by-step explanation:
An uniform probability is a case of probability in which each outcome is equally as likely.
For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.
The probability that we find a value X lower than x is given by the following formula.
![P(X \leq x) = \frac{x - a}{b-a}](https://tex.z-dn.net/?f=P%28X%20%5Cleq%20x%29%20%3D%20%5Cfrac%7Bx%20-%20a%7D%7Bb-a%7D)
For this problem, we have that:
Uniformly distributed between 3 minutes and 15 minutes:
So ![a = 3, b = 15](https://tex.z-dn.net/?f=a%20%3D%203%2C%20b%20%3D%2015)
What is the probability that it takes Isabella more than 11 minutes to wait for the bus?
Either she has to wait 11 or less minutes for the bus, or she has to wait more than 11 minutes. The sum of these probabilities is 1. So
![P(X \leq 11) + P(X > 11) = 1](https://tex.z-dn.net/?f=P%28X%20%5Cleq%2011%29%20%2B%20P%28X%20%3E%2011%29%20%3D%201)
We want P(X > 11). So
![P(X > 11) = 1 - P(X \leq 11) = 1 - \frac{11 - 3}{15 - 3} = 0.3333](https://tex.z-dn.net/?f=P%28X%20%3E%2011%29%20%3D%201%20-%20P%28X%20%5Cleq%2011%29%20%3D%201%20-%20%5Cfrac%7B11%20-%203%7D%7B15%20-%203%7D%20%3D%200.3333)
33.33% probability that it takes Isabella more than 11 minutes to wait for the bus
800 * 100 = 80000
810 * 99 = 80190
900 * 90 = 81000
1000 * 80 = 80000
(800 + 10n) (100-1n)
=(800 +10(11)) (100 - 11)<span>
(910)(89) = 80990</span>
It is 77.5. Using a calculator is more useful. ☺
The answer:
![\sqrt{9}1 /10](https://tex.z-dn.net/?f=%5Csqrt%7B9%7D1%20%2F10)
Explanation to your question:
Since the sin of theta is 0.3, we can reasonably deduct that the opposite side to theta has a ration of 3 to 10 to that of the hypotenuse. Thus, the adjacent side to theta, using the pythagorean theorem, will be root91. Therefore, since the cosine of theta is the adjacent/hypotenuse, we get root 91/10
Answer:
2^3 or 8
Step-by-step explanation:
![4^3\div 2^3= \\\\(2^2)^3\div 2^3= \\\\2^{2\cdot 3}\div 2^3= \\\\2^6\div 2^3=\\\\2^{6-3}=\\\\2^3=\\\\8](https://tex.z-dn.net/?f=4%5E3%5Cdiv%202%5E3%3D%20%5C%5C%5C%5C%282%5E2%29%5E3%5Cdiv%202%5E3%3D%20%5C%5C%5C%5C2%5E%7B2%5Ccdot%203%7D%5Cdiv%202%5E3%3D%20%5C%5C%5C%5C2%5E6%5Cdiv%202%5E3%3D%5C%5C%5C%5C2%5E%7B6-3%7D%3D%5C%5C%5C%5C2%5E3%3D%5C%5C%5C%5C8)
Hope this helps!