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Alik [6]
4 years ago
8

If jobs arrive every 15 seconds on average, what is the probability of waiting more than 30 seconds?

Mathematics
1 answer:
Aleksandr [31]4 years ago
5 0

Answer: 0.14

Step-by-step explanation:

Given: Mean : \lambda=15\text{ per seconds}

In minutes , Mean : \lambda=4\text{ per minute}

The exponential distribution function with parameter \lambda  is given by :-

f(t)=\lambda e^{-\lambda t}, \text{ for }x\geq0

The probability of waiting more than 30 seconds i.e. 0.5 minutes is given by the exponential function :-

P(X\geq0.5)=1-P(X\leq0.5)\\\\=1-\int^{0.5}_{0}4e^{-4t}dt\\\\=1-[-e^{-4t}]^{0.5}_{0}\\\\=1-(1-e^{-2})=1-0.86=0.14

Hence, the probability of waiting more than 30 seconds = 0.14

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Kendrick adds 1/2cup of chicken stock to a pot. Then he takes 1/2 cup of stock out of the pot. What is the overall increase or d
Hoochie [10]

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3 years ago
Can you help me pls, this doesn't make any sense
Gala2k [10]
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4 years ago
How To Solve these? ​
Inessa [10]

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b. Same, when the denominator is the same, you can just minus the numerators. Which becomes, 25-12=13--> 13/27

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5 0
3 years ago
A+b+c=31 2a+3b+4c=105 who's a,b,c
Alik [6]

Answer:

begin{aligned}

   & x = \frac{D_x}{D} = \frac{-616}{-154} =  4 \\

   & y = \frac{D_y}{D} = \frac{ 616}{-154} = -4 \\

   & z = \frac{D_z}{D} = \frac{-770}{-154} =  5

         \end{aligned}begin{aligned}    & x = \frac{D_x}{D} = \frac{-616}{-154} =  4 \\    & y = \frac{D_y}{D} = \frac{ 616}{-154} = -4 \\    & z = \frac{D_z}{D} = \frac{-770}{-154} =  5          \end{aligned}

Step-by-step explanation:

4 0
4 years ago
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