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3 years ago

Height, in meters, is measured for each person in a sample. After the data are collected, all the height measurements

Mathematics

1 answer:

charle [14.2K]3 years ago

5 0

Answer:

(E) The z-scores of the height measurements

Step-by-step explanation:

Mean is given by

$\bar {x}=\frac{1}{n}\left(\sum _{i=1}^{n}{x_{i}}\right)=\frac{x_{1}+x_{2}+\cdots +x_{n}}{n}$

Where,

n = Number of people

$x_{i}$ = The heights of people

When converting m to cm the mean would be

$\bar {x}=\frac{1}{n}\left(\sum _{i=1}^{n}{x_{i}}\right)\times 100$

So, the mean would change

The median gives us the middle data value when the data values are in ascending order.

So, the median would change

Standard deviation

$s=\sqrt{\frac{1}{N-1}\sum_{i=1}^{N}(x_i-\bar{x})^2}$

When converting to centimeters

$s=\sqrt{\frac{1}{N-1}\sum_{i=1}^{N}\left((x_i-\bar{x})\times 100\right)^2$

Hence, the standard deviation would change

When converting to centimeters the maximum height in meters would be the maximum height in centimeters also.

The z score is given by

$z=\frac{x-\mu}{\sigma}$

where,

x = Data point

$\mu$ = Mean

$\sigma$ = Standard deviation

When converting to centimeters

$z=\frac{(x-\mu)\times 100}{\sigma\times 100}\\\Rightarrow z=\frac{x-\mu}{\sigma}$

Hence, the z score would remain the same

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3 years ago

A 1000-liter (L) tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flo

djverab [1.8K]

Answer:

a) $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) $31690.7 g/L$

Step-by-step explanation:

By definition, we have that the change rate of salt in the tank is $\frac{dy}{dt}=R_{i}-R_{o}$ , where $R_{i}$ is the rate of salt entering and $R_{o}$ is the rate of salt going outside.

Then we have, $R_{i}=80\frac{L}{min}*50\frac{g}{L}=4000\frac{g}{min}$ , and

$R_{o}=40\frac{L}{min}*\frac{y}{500} \frac{g}{L}=\frac{2y}{25}\frac{g}{min}$

So we obtain. $\frac{dy}{dt}=4000-\frac{2y}{25}$ , then

$\frac{dy}{dt}+\frac{2y}{25}=4000$ , and using the integrating factor $e^{\int {\frac{2}{25}} \, dt=e^{\frac{2t}{25}$ , therefore $(\frac{dy }{dt}+\frac{2y}{25}}=4000)e^{\frac{2t}{25}$ , we get $\frac{d}{dt}(y*e^{\frac{2t}{25}})= 4000 e^{\frac{2t}{25}$ , after integrating both sides $y*e^{\frac{2t}{25}}= 50000 e^{\frac{2t}{25}}+C$ , therefore $y(t)= 50000 +Ce^{\frac{-2t}{25}}$ , to find $C$ we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions $y(0)=10$ , so $10= 50000+Ce^{\frac{-0*2}{25}}$

$10=50000+C\\C=10-50000=-49990$

Finally we can write an expression for the amount of salt in the tank at any time t, it is $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) The tank will overflow due Rin>Rout, at a rate of $80 L/min-40L/min=40L/min$ , due we have 500 L to overflow $\frac{500L}{40L/min} =\frac{25}{2} min=t$ , so we can evualuate the expression of a) $y(25/2)=50000-49990e^{\frac{-2}{25}\frac{25}{2}}=50000-49990e^{-1}=31690.7$ , is the salt concentration when the tank overflows