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3 years ago

Rearrange the formula A = Θr2 2 for Θ.

Mathematics

2 answers:

In-s [12.5K]3 years ago

5 0

Answer:

$\theta=\frac{2A}{r^{2} }$

Step-by-step explanation:

The given formula is

$A=\frac{\theta r^{2} }{2}$

We have to solve the formula for $\theta$ . To do that, we first need to past the divisor 2 to the other side, as follows

$2A=\theta r^{2}$

Then, we have to move the squared radius to the other side with the opposite operation, that is, dividing

$\frac{2A}{r^{2} } =\theta \\\theta=\frac{2A}{r^{2} }$

Threfore, the given expression solved for $\theta$ is

$\theta=\frac{2A}{r^{2} }$

Lina20 [59]3 years ago

3 0

Answer:

$\theta=\frac{2A}{r^2}$

Step-by-step explanation:

Given : The formula $A=\frac{\theta r^2}{2}$

We have to rearrange the given formula for $\theta$

Consider the given formula $A=\frac{\theta r^2}{2}$

Multiply both side by 2, we have,

$2A={\theta r^2$

Divide both side by $r^2$ , we have,

$\frac{2A}{r^2}= \frac{\theta r^2}{r^2}$

Simplify, we get,

$\frac{2A}{r^2}=\theta$

Thus, $\theta=\frac{2A}{r^2}$

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3 years ago

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2 years ago

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3 years ago

SAT scores are normed so that, in any year, the mean of the verbal or math test should be 500 and the standard deviation 100. as

vovangra [49]

Answer:

a) $P(X>625)=P(\frac{X-\mu}{\sigma}>\frac{625-\mu}{\sigma})=P(Z>\frac{625-500}{100})=P(Z>1.25)$

$P(Z>1.25)=1-P(Z$

b) $P(400$

$P(-1$

c) $z=-0.842$

And if we solve for a we got

$a=500 -0.842*100=415.8$

So the value of height that separates the bottom 20% of data from the top 80% is 415.8.

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".

Part a

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

$X \sim N(500,100)$

Where $\mu=500$ and $\sigma=100$

We are interested on this probability

$P(X>625)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we apply this formula to our probability we got this:

$P(X>625)=P(\frac{X-\mu}{\sigma}>\frac{625-\mu}{\sigma})=P(Z>\frac{625-500}{100})=P(Z>1.25)$

And we can find this probability using the complement rule and with the normal standard table or excel:

$P(Z>1.25)=1-P(Z$

Part b

We are interested on this probability

$P(400$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we apply this formula to our probability we got this:

$P(400$

And we can find this probability with this difference:

$P(-1$

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.

$P(-1$

Part c

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.8$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.2 of the area on the left and 0.8 of the area on the right it's z=-0.842. On this case P(Z<-0.842)=0.2 and P(Z>-0.842)=0.8

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=-0.842$

And if we solve for a we got

$a=500 -0.842*100=415.8$

So the value of height that separates the bottom 20% of data from the top 80% is 415.8.

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2 years ago