I need more help =) ;) :p

Consider the polynomial function g(x)=5x^6+x^5+9x^3-12x-125 What is the end behavior of the graph of g?

Slav-nsk [51]

Answer:

As x —> infinity, g(x)—> infinity, and as x —> Negative infinity, g(x) —> Infinity

Step-by-step explanation:

Because

7 0

3 years ago

Solve by quadratic formula

aliya0001 [1]

8). x=(-1+√22)/2 , x=(-1-√22)/3

9). x=(1+√5)/2 , x=(1-√5)/2

10). x=3 , 1/4

11). x=1/3 , -1

12). x=2+2√3 , 2-2√3

4 0

3 years ago

Write y=5x-9 in standard form<br> Will give brainliest!

ArbitrLikvidat [17]

Answer:

5x - y = 9

Step-by-step explanation:

the equation of a line in standard form is

Ax + By = C ( A is a positive integer and B, C are integers )

given

y = 5x - 9 ( subtract 5x from both sides )

- 5x + y = - 9 ( multiply through by - 1 )

5x - y = 9 ← in standard form

6 0

2 years ago

Graph. Y= 2/3x - 3<br> Pls need help will very apperciate

zhuklara [117]

Answer:

Here is what the graph will look like and here are some points..

4 0

3 years ago

Assume that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true

Feliz [49]

Answer:

a) The 95% CI for the true average porosity is (4.51, 5.19).

b) The 98% CI for true average porosity is (4.11, 5.01)

c) A sample size of 15 is needed.

d) A sample size of 101 is needed.

Step-by-step explanation:

a. Compute a 95% CI for the true average porosity of a certain seam if the average porosity for 20 specimens from the seam was 4.85.

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96*\frac{0.78}{\sqrt{20}} = 0.34$

The lower end of the interval is the sample mean subtracted by M. So it is 4.85 - 0.34 = 4.51

The upper end of the interval is the sample mean added to M. So it is 4.35 + 0.34 = 5.19

The 95% CI for the true average porosity is (4.51, 5.19).

b. Compute a 98% CI for true average porosity of another seam based on 16 specimens with a sample average of 4.56.

Following the same logic as a.

98% C.I., so $z = 2.327$

$M = 2.327*\frac{0.78}{\sqrt{16}} = 0.45$

4.56 - 0.45 = 4.11

4.56 + 0.45 = 5.01

The 98% CI for true average porosity is (4.11, 5.01)

c. How large a sample size is necessary if the width of the 95% interval is to be 0.40?

A sample size of n is needed.

n is found when M = 0.4.

95% C.I., so Z = 1.96.

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.4 = 1.96*\frac{0.78}{\sqrt{n}}$

$0.4\sqrt{n} = 1.96*0.78$

$\sqrt{n} = \frac{1.96*0.78}{0.4}$

$(\sqrt{n})^{2} = (\frac{1.96*0.78}{0.4})^{2}$

$n = 14.6$

Rounding up

A sample size of 15 is needed.

d. What sample size is necessary to estimate the true average porosity to within 0.2 with 99% confidence?

99% C.I., so z = 2.575

n when M = 0.2.

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.2 = 2.575*\frac{0.78}{\sqrt{n}}$

$0.2\sqrt{n} = 2.575*0.78$

$\sqrt{n} = \frac{2.575*0.78}{0.2}$

$(\sqrt{n})^{2} = (\frac{2.575*0.78}{0.2})^{2}$

$n = 100.85$

Rounding up

A sample size of 101 is needed.

8 0

4 years ago