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andreev551 [17]
3 years ago
13

F granola costs $2.20 per pound, how much does 0.8 pound of granola cost?

Mathematics
2 answers:
Varvara68 [4.7K]3 years ago
4 0

Set up the proportion.  Then cross multiply to solve for the variable.

\frac{2.20 dollars}{1 pound} = \frac{x}{0.8 pounds}

2.20(0.8) = 1(x)

1.76 = x

Answer: $1.76


ki77a [65]3 years ago
3 0

Answer: 1.76 $

2.20 times 0.8 is 1.76

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10³/10³

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hope it helps..

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Answer:

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d) L_q =\frac{20^2}{30(30-20)}=1.333 people

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Step-by-step explanation:

Notation

P represent the probability that the employee is idle

p_x represent the probability that the employee is busy

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W_s represent the average time a person seeking information spends in the system

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This an special case of Single channel model

Single Channel Queuing Model. "That division of service channels happen in regards to number of servers that are present at each of the queues that are formed. Poisson distribution determines the number of arrivals on a per unit time basis, where mean arrival rate is denoted by λ".

Part a

Find the probability that the employee is idle

The probability on this case is given by:

In order to find the mean we can do this:

\mu = \frac{1question}{2minutes}\frac{60minutes}{1hr}=\frac{30 question}{hr}

And in order to find the probability we can do this:

P=1-\frac{\lambda}{\mu}=1-\frac{20}{30}=0.33 and that represent the 33%

Part b

Find the proportion of the time that the employee is busy

This proportion is given by:

p_x =\frac{\lambda}{\mu}=\frac{20}{30}=0.66

Part c

Find the average number of people receiving and waiting to receive some information

In order to find this average we can use this formula:

L_s= \frac{\lambda}{\lambda -\mu}

And replacing we got:

L_s =\frac{20}{30-20}=\frac{20}{10}=2 people

Part d

Find the average number of people waiting in line to get some information.

For the number of people wiating we can us ethe following formula"

L_q =\frac{\lambda^2}{\mu(\mu-\lambda)}

And replacing we got this:

L_q =\frac{20^2}{30(30-20)}=1.333 people

Part e

Find the average time a person seeking information spends in the system

For this average we can use the following formula:

W_s =\frac{1}{\lambda -\mu}=\frac{1}{30-20}=0.1hours

Part f

Find the expected time a person spends just waiting in line to have a question answered (time in the queue).

For this case the waiting time to answer a question we can use this formula:

W_q =\frac{\lambda}{\mu(\mu -\lambda)}=\frac{20}{30(30-20)}=0.0667 hours

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