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Anon25 [30]
3 years ago
13

(x - 1) {}^{2} - 3" alt="g(x) = (x - 1) {}^{2} - 3" align="absmiddle" class="latex-formula">
Find...
g {}^{2} (1)
​
Mathematics
1 answer:
I am Lyosha [343]3 years ago
3 0

Answer:

9.

Step-by-step explanation

g(x) = (x - 1)^2 - 3

g(1) = (1-1)^2 - 3 = -3

so g^2 (1) = (-3) ^2

= 9.

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3 to the power of 5 = 35 = 243

Step-by-step explanation:

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What is the derivative of w=(t^2+1)^100<br><br> How is it 200t(t^2+1)^99?
GuDViN [60]
(t^2+1)^100

USE CHAIN RULE

Outside first (using power rule)

100*(t^2+1)^99 * derivative of the inside

100(t^2+1)^99 * d(t^2+1)

100(t^2+1)^99 * 2t

200t(t^2+1)^99
6 0
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A forester studying the effects of fertilization on certain pine forests in the Southeast is interested in estimating the averag
ahrayia [7]

Answer:

P(-2 \leq \bar X -\mu \leq 2)

If we divide both sides by \frac{\sigma}{\sqrt{n}} we got:

P(\frac{-2}{\frac{4}{\sqrt{9}}}\leq Z \leq \frac{2}{\frac{4}{\sqrt{9}}})

And we can use the normal distribution table or excel to find the probabilites and we got:

P(-1.5 \leq Z \leq 1.5)= P(ZStep-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 area of a population, and for this case we know the distribution for X is given by:

X \sim N(M,4)  

Where \mu=M and \sigma=4

We select a a sample of n =4 and since the distribution for X is normal then we know that the distribution for the sample mean \bar X is given by:

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

And we want to find this probability:

P(-2 \leq \bar X -\mu \leq 2)

If we divide both sides by \frac{\sigma}{\sqrt{n}} we got:

P(\frac{-2}{\frac{4}{\sqrt{9}}}\leq Z \leq \frac{2}{\frac{4}{\sqrt{9}}})

And we can use the normal distribution table or excel to find the probabilites and we got:

P(-1.5 \leq Z \leq 1.5)= P(Z

8 0
3 years ago
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