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bogdanovich [222]
3 years ago
8

The number of boxes in Warehouse "A" currently totaling 476 is increasing at a rate of 4 boxes per day. The number of boxes in W

arehouse "B" currently totaling 986 is decreasing at a rate of 6 boxes per day. In how many days will these two warehouses contain the same number of boxes
Mathematics
1 answer:
Bingel [31]3 years ago
3 0

Answer:

These two warehouses will contain the same number of boxes in 51 days.

Step-by-step explanation:

Since the number of boxes in Warehouse "A" currently totaling 476 is increasing at a rate of 4 boxes per day, while the number of boxes in Warehouse "B" currently totaling 986 is decreasing at a rate of 6 boxes per day, To determine in how many days these two warehouses contain the same number of boxes, the following calculation must be performed:

Warehouse A = 476 + 4X

Warehouse B = 986 - 6X

476 + 4 x 50 = 476 + 200 = 676

986 - 6 x 50 = 986 - 300 = 686

476 + 4 x 51 = 476 + 204 = 680

986 - 6 x 51 = 986 - 306 = 680

Therefore, these two warehouses will contain the same number of boxes in 51 days.

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The time between telephone calls to a cable television service call center follows an exponential distribution with a mean of 1.
Ulleksa [173]

Answer:

0.52763 is the probability that the time between the next two calls will be 54 seconds or​ less.

0.19285 is the probability that the time between the next two calls will be greater than 118.5 ​seconds.

Step-by-step explanation:

We are given the following information in the question:

The time between telephone calls to a cable television service call center follows an exponential distribution with a mean of 1.2 minutes.

The distribution function can be written as:

f(x) = \lambda e^{-\lambda x}\\\text{where lambda is the parameter}\\\\\text{Mean} = \mu = \displaystyle\frac{1}{\lambda}\\\\\Rightarrow 1.2 = \frac{1}{\lambda}\\\\\lambda = 0.84 \\f(x) = 0.84 e^{0.84 x}

The probability for exponential distribution is given as:

P( x \leq a) = 1 - e^{\frac{-a}{\mu}}\\\\P(a \leq x \leq b) = e^{\frac{-a}{\mu} -\frac{-b}{\mu}}

a) P( time between the next two calls will be 54 seconds or​ less)

P( x \leq 0.9)\\= 1 - e^{\frac{\frac{-54}{60}}{1.2}} = 0.52763

0.52763 is the probability that the time between the next two calls will be 54 seconds or​ less.

b) P(time between the next two calls will be greater than 118.5 ​seconds)

p( x > \frac{118.5}{60}) = P(x > 1.975)\\\\ = 1 - P(x \leq 1.975) \\\\= 1 -1+ e^{\frac{-1.975}{1.2}}\\\\= 0.19285

0.19285 is the probability that the time between the next two calls will be greater than 118.5 ​seconds.

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Step-by-step explanation:

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