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ollegr [7]
3 years ago
13

Use the following cell phone airport data speeds​ (Mbps) from a particular network. Find the percentile corresponding to the dat

a speed 4.9 Mbps.
0.2 0.8 2.3 6.4 12.3 0.2 0.8 2.3 6.9 12.7 0.2 0.8 2.6 7.5 12.9 0.3 0.9 2.8 7.9 13.8
0.6 1.5 0.1 0.7 2.2 6.1 12.1 0.6 1.9 5.5 11.9 27.5 0.6 1.7 3.3 8.3 13.8 1.3 3.5 9.8
14.6 10.1 14.7 11.8 14.8
Mathematics
1 answer:
Tems11 [23]3 years ago
7 0

Answer:

Thus percentile lies between 53.3% and 55.6 %

Step-by-step explanation:

First we arrange the data in ascending order . Then find the number of the values corresponding to the given value. Then equate it with the number of observations and x and then multiply it to get the percentile. n= P/100 *N

where n is the ordinal rank of the given value

N is the number of values in ascending order.

The data in ascending order is

0.1 0.2 0.2 0.2 0.3 0.6 0.6 0.6 0.7 0.8 0.8 0.8 0.9 1.3

1.5 1.7 1.9 2.2 2.3 2.3 2.6 2.8 3.3 3.5 5.5 6.1 6.4 6.9 7.5 7.9 8.3 9.8 10.1 11.8 11.9 12.1 12.3 12.7 12.9 13.8 13.8 14.6 14.7 14.8 27.5

Number of observation = 45

4.9 lies between 3.3 and 5.5

x*n = 24 observation x*n = 25 observation

x*45= 24 x*45= 25

x= 0.533 x= 0.556

Thus percentile lies between 53.3% and 55.6 %

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Answer:

The quantity of salt at time t is m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} }), where t is measured in minutes.

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\dot m_{in} - \dot m_{out} = \left(\frac{dm}{dt} \right)_{CV}

Where mass flow is the product of salt concentration and water volume flow.

The model of the tank according to the statement is:

(0.5\,\frac{pd}{gal} )\cdot \left(6\,\frac{gal}{min} \right) - c\cdot \left(6\,\frac{gal}{min} \right) = V\cdot \frac{dc}{dt}

Where:

c - The salt concentration in the tank, as well at the exit of the tank, measured in \frac{pd}{gal}.

\frac{dc}{dt} - Concentration rate of change in the tank, measured in \frac{pd}{min}.

V - Volume of the tank, measured in gallons.

The following first-order linear non-homogeneous differential equation is found:

V \cdot \frac{dc}{dt} + 6\cdot c = 3

60\cdot \frac{dc}{dt}  + 6\cdot c = 3

\frac{dc}{dt} + \frac{1}{10}\cdot c = 3

This equation is solved as follows:

e^{\frac{t}{10} }\cdot \left(\frac{dc}{dt} +\frac{1}{10} \cdot c \right) = 3 \cdot e^{\frac{t}{10} }

\frac{d}{dt}\left(e^{\frac{t}{10}}\cdot c\right) = 3\cdot e^{\frac{t}{10} }

e^{\frac{t}{10} }\cdot c = 3 \cdot \int {e^{\frac{t}{10} }} \, dt

e^{\frac{t}{10} }\cdot c = 30\cdot e^{\frac{t}{10} } + C

c = 30 + C\cdot e^{-\frac{t}{10} }

The initial concentration in the tank is:

c_{o} = \frac{10\,pd}{60\,gal}

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0.167 = 30 + C

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The solution of the differential equation is:

c(t) = 30 - 29.833\cdot e^{-\frac{t}{10} }

Now, the quantity of salt at time t is:

m_{salt} = V_{tank}\cdot c(t)

m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })

Where t is measured in minutes.

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