1. <span><span>20,000 </span><span>+7,000 </span><span>+500 </span><span>+40 </span><span>+9 -----WORD FORM :</span></span>twenty-seven thousand,
five hundred forty-nine
2. <span>Expanded Numbers Form:
</span>
<span><span> 700,000 </span><span>+90,000 </span><span>+2,000 </span><span>+0 </span><span>+60 </span><span>+5 </span></span>
WORD FORM : seven hundred ninety-two thousand,
sixty-five
Answer: The three different equations will be

Step-by-step explanation:
We have to write three equation that have x = 5 as a solution:
1) First equation will be

2) Second equation will be

3) Third equation will be

Hence, the three different equations will be

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:

a) The probability that exactly 3 arrivals occur during a particular hour is:

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:

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.

-4.2, -3.5,-2.1,-1.5,-1,-0.5,0.5,2,3.5,4.8
Answer:
Step-by-step explanation: