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3 years ago

Standard deviation of a normal data distribution is a _______. Group of answer choices

Mathematics

1 answer:

FrozenT [24]3 years ago

5 0

Answer:

Standard deviation of a normal data distribution is a measure of data dispersion.

Step-by-step explanation:

Standard deviation is used to measure dispersion which is present around the mean data.

The value of standard deviation will never be negative.

The greater the spread, the greater the standard deviation.

Steps-

1. At first, the mean value should be discovered.

2.Then find out the square of it's distance to mean value.

3.Then total the values

4.Then divide the number of data point.

5.the square root have to be taken.

Formula-

SD= $\sqrt{\frac{(\sum{x-x)^2} }{n-1}$

Advantage-

It is used to measure dispersion when mean is used as measure of central tendency.

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A ball is thrown into the air and its position is given by h ( t ) = − 6.3 t 2 + 53 t + 24 where h is the height of the ball in

nordsb [41]

Answer:

$h'(t) = -12.6 t +53$

Now we can set up the derivate equal to 0 and we have:

$-12.6 t +53 = 0$

And solving for t we got:

$t = \frac{53}{12.6}= 4.206$

For the second derivate respect the time we got:

$h''(t) = -12.6$

So then we can conclude that t = 4.206 is a maximum for the function.

And the corresponding height would be:

$h(t=4.206) = -6.3(4.206)^2 +53*4.206+24= 135.468 ft$

So the maximum occurs at t = 4.206 s and with a height of 135.468 m

Step-by-step explanation:

For this case we have the following function:

$h(t) = -6.3t^2 +53 t+24$

In order to maximize this function we need to take the first derivate respect the time and we have:

$h'(t) = -12.6 t +53$

Now we can set up the derivate equal to 0 and we have:

$-12.6 t +53 = 0$

And solving for t we got:

$t = \frac{53}{12.6}= 4.206$

For the second derivate respect the time we got:

$h''(t) = -12.6$

So then we can conclude that t = 4.206 is a maximum for the function.

And the corresponding height would be:

$h(t=4.206) = -6.3(4.206)^2 +53*4.206+24= 135.468 ft$

So the maximum occurs at t = 4.206 s and with a height of 135.468 m

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