I need help! plsssssssssssssssss

Which is the graph of g(x) = [X +3]? 32-1.

raketka [301]

Answer:

44. f(x) = x/(x2 * 1) 1 45. f(x) = 3x15 * 5x3 + 3 46. f(x) = x + — x 47. (a) Use a graphing calculator or computer to graph the function f(x) = x4 * 3x3 * 6x2 + 7x + 30 in the viewing 1 l3. F(x)=(%x)5 ... t 3|. '=* 32. = > 1) y (t*1)2 t2+2 r=\/Y 33. '= 34.

Step-by-step explanation:

4 0

3 years ago

A company interested in lumbering rights for a certain tract of slash pine trees is told that the mean diameter of these trees i

brilliants [131]

Answer:

97.6% of the trees will have diameters between 7.6 and 18.6 inches.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 12 inches

Standard Deviation, σ = 2.2 inches

We are given that the distribution of diameter of tree is a mound shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(diameters between 7.6 and 18.6 inches)

$P(7.6 \leq x \leq 18.6) = P(\displaystyle\frac{7.6 - 12}{2.2} \leq z \leq \displaystyle\frac{18.6-12}{2.2}) = P(-2 \leq z \leq 3)\\\\= P(z \leq 3) - P(z < -2)\\= 0.999 - 0.023 = 0.976= 97.6\%$

$P(7.6 \leq x \leq 18.6) = 97.6\%$

97.6% of the trees will have diameters between 7.6 and 18.6 inches.

8 0

3 years ago

In a randomly selected sample of 100 students at a University, 81 of them had access to a computer at home. Give the value of th

astra-53 [7]

Answer:

The value of the standard error for the point estimate is of 0.0392.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$