Answer:
33.3333...
Step-by-step explanation:
50000 ÷ 15000, which is the increase = 0.3333333
Times by 100 for percent
The area is 2,772 feet
Hope this helps! Good luck on your homework!!!
Answer:
Choice A. 3.
Step-by-step explanation:
The triangle in question is a right triangle.
- The length of the hypotenuse (the side opposite to the right angle) is given.
- The measure of one of the acute angle is also given.
As a result, the length of both legs can be found directly using the sine function and the cosine function.
Let
denotes the length of the side opposite to the
acute angle, and
be the length of the side next to this
acute angle.
.
Similarly,
.
The longer leg in this case is the one adjacent to the
acute angle. The answer will be
.
There's a shortcut to the answer. Notice that
. The cosine of an acute angle is directly related to the adjacent leg. In other words, the leg adjacent to the
angle will be the longer leg. There will be no need to find the length of the opposite leg.
Does this relationship
holds for all acute angles? (That is,
?) It turns out that:
Answer:
a) P(x=3)=0.089
b) P(x≥3)=0.938
c) 1.5 arrivals
Step-by-step explanation:
Let t be the time (in hours), then random variable X is the number of people arriving for treatment at an emergency room.
The variable X is modeled by a Poisson process with a rate parameter of λ=6.
The probability of exactly k arrivals in a particular hour can be written as:
![P(x=k)=\lambda^{k} \cdot e^{-\lambda}/k!\\\\P(x=k)=6^k\cdot e^{-6}/k!](https://tex.z-dn.net/?f=P%28x%3Dk%29%3D%5Clambda%5E%7Bk%7D%20%5Ccdot%20e%5E%7B-%5Clambda%7D%2Fk%21%5C%5C%5C%5CP%28x%3Dk%29%3D6%5Ek%5Ccdot%20e%5E%7B-6%7D%2Fk%21)
a) The probability that exactly 3 arrivals occur during a particular hour is:
![P(x=3)=6^{3} \cdot e^{-6}/3!=216*0.0025/6=0.089\\\\](https://tex.z-dn.net/?f=P%28x%3D3%29%3D6%5E%7B3%7D%20%5Ccdot%20e%5E%7B-6%7D%2F3%21%3D216%2A0.0025%2F6%3D0.089%5C%5C%5C%5C)
b) The probability that <em>at least</em> 3 people arrive during a particular hour is:
![P(x\geq3)=1-[P(x=0)+P(x=1)+P(x=2)]\\\\\\P(0)=6^{0} \cdot e^{-6}/0!=1*0.0025/1=0.002\\\\P(1)=6^{1} \cdot e^{-6}/1!=6*0.0025/1=0.015\\\\P(2)=6^{2} \cdot e^{-6}/2!=36*0.0025/2=0.045\\\\\\P(x\geq3)=1-[0.002+0.015+0.045]=1-0.062=0.938](https://tex.z-dn.net/?f=P%28x%5Cgeq3%29%3D1-%5BP%28x%3D0%29%2BP%28x%3D1%29%2BP%28x%3D2%29%5D%5C%5C%5C%5C%5C%5CP%280%29%3D6%5E%7B0%7D%20%5Ccdot%20e%5E%7B-6%7D%2F0%21%3D1%2A0.0025%2F1%3D0.002%5C%5C%5C%5CP%281%29%3D6%5E%7B1%7D%20%5Ccdot%20e%5E%7B-6%7D%2F1%21%3D6%2A0.0025%2F1%3D0.015%5C%5C%5C%5CP%282%29%3D6%5E%7B2%7D%20%5Ccdot%20e%5E%7B-6%7D%2F2%21%3D36%2A0.0025%2F2%3D0.045%5C%5C%5C%5C%5C%5CP%28x%5Cgeq3%29%3D1-%5B0.002%2B0.015%2B0.045%5D%3D1-0.062%3D0.938)
c) In this case, t=0.25, so we recalculate the parameter as:
![\lambda =r\cdot t=6\;h^{-1}\cdot 0.25 h=1.5](https://tex.z-dn.net/?f=%5Clambda%20%3Dr%5Ccdot%20t%3D6%5C%3Bh%5E%7B-1%7D%5Ccdot%200.25%20h%3D1.5)
The expected value for a Poisson distribution is equal to its parameter λ, so in this case we expect 1.5 arrivals in a period of 15 minutes.
![E(x)=\lambda=1.5](https://tex.z-dn.net/?f=E%28x%29%3D%5Clambda%3D1.5)
Answer: 21 is your range
hope this helps
Step-by-step explanation:
To find the range, subtract the smallest number in the set from the largest