Answer:
The answer to this equation will be 2 for x
Step-by-step explanation:
Hope this Helped
Answer: D. 7
Step-by-step explanation:
The problem tells you that x = 0 so replace everything that is x in the equation with 0 so the problem becomes
7 + (-3(0)^2)
Now we just solve the problem normally.
solve the problem in the parenthesis first, we must multiply -3 times 0^2 is 0
so now our problem is 7 + 0 which is 7
Answer:
Para trasladar 300 cajones, las dos opciones cuestan los mismo. Son indiferentes entre si.
Step-by-step explanation:
<u>Dada la siguiente información:</u>
<u></u>
Cantidad de cajones= 300
Opción A:
Camiones con capacidad para 50 cajones que cobran 150 por viaje.
Cantidad de camiones= 300/50= 6
Opción B:
Camiones con capacidad de 60 cajones por 180 por viaje.
Cantidad de camiones= 300/60= 5
<u>Debemos calcular el costo total de cada opción y elegir la de menor costo.</u>
Costo total A= 150*6= $900
Costo total B=180*5= $900
Para trasladar 300 cajones, las dos opciones cuestan los mismo. Son indiferentes entre si.
X= number of adult tickets
y=number of child tickets.
we can suggest this system of equations:
x+y=118 ⇒x=118-y
7.5x+3y=696
we can solve this system of equations by substitution method:
7.5(118-y)+3y=696
885-7.5y+3y=696
-4.5y=-189
y=-189/-4.5
y=42
x=118-y
x=118-42
x=76
Answer: the number of adult tickets sold was 76.
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.
