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Novay_Z [31]
3 years ago
5

Solve for the indicated variable. a=f/m for m SHOW ALL WORK

Mathematics
1 answer:
Lena [83]3 years ago
3 0
Ok, a=f/m
What we need to do is isolate the variable, m.
  a=f/m
*m  *m
a*m=f
m*a=f
/a     /a
m=f/a
There's your answer.
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Finding Derivatives Implicity In Exercise, find dy/dx implicity.<br> x2 - 3 ln y + y2 = 10
Veseljchak [2.6K]

Answer:

\frac{dy}{dx}=-\frac{2xy}{2y^2-3}

Step-by-step explanation:

We are given that

x^2-3lny+y^2=0

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2x-\frac{3}{y}\frac{dy}{dx}+2y\frac{dy}{dx}=0

By using formula

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\frac{dy^n}{dx}=ny^{n-1}\frac{dy}{dx}

\frac{dy}{dx}(-\frac{3}{y}+2y)+2x=0

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Hence, the derivative of function

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8 0
3 years ago
What is the mean, medium, mode of 54,220
irina [24]

Answer:

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Median: 15.5

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Range: 16

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Step-by-step explanation:

hope this helps!

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2 years ago
Intravenous fluid bags are filled by an automated filling machine. Assume that the fill volumes of the bags are independent, nor
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Answer:

Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:

X \sim N(\mu,0.08)  

Where \mu and \sigma=0.08

Since the distribution for X is normal then the  we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And the standard error is given by:

SE= \frac{0.08}{\sqrt{24}}= 0.0163

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:

X \sim N(\mu,0.08)  

Where \mu and \sigma=0.08

Since the distribution for X is normal then the  we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And the standard error is given by:

SE= \frac{0.08}{\sqrt{24}}= 0.0163

8 0
3 years ago
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