All categories

3 years ago

Helppppppp pleaseeeee

Mathematics

1 answer:

ratelena [41]3 years ago

4 0

Let's consider each of the options;

A) The range is the values that y can be of the function f(x). We can see that no matter what x values you put in, f(x)>0. It will never be negative. Although, if you put (1/2)^2, you can get a y value of (1/4) so this statement is incorrect.

B) It would have to put in the x or y value into the function and check. f(0)=(1/2)^0

You know that anything to the zero-th power gives 1, therefore, this statement is correct.

C) Again, if you put in a value such as (1/2)^2, you would be getting a number that is smaller than 1/2, so it isn't always increasing.

D) Again, it is possible to input any value in x and you would still be getting a positive y-value, therefore, this statement is incorrect.

Hope I helped :)

You might be interested in

Alecia builds a box in the shape of a cube with an open top. She plans to paint the box. Alecia calculates that there are 256 sq

STALIN [3.7K]

Answer:

See Explanation

Step-by-step explanation:

The question is incomplete, as the options to select the required condition from were not listed. So, I will answer on general terms

From the question, we understand that the cube has an open-top and the surface ares is 256 in^2

Let L represents the edge length, the surface area (S) is:

$S = 5L^2$

Substitute 256 for S

$256 = 5L^2$

Divide both sides by 5

$51.2 = L^2$

Take square roots of both sides:

$\sqrt{51.2} = L$

$7.155= L$

$L=7.16in$ ---- approximated

The volume of the cube is:

$Volume = L^3$

$Volume = \sqrt{51.2}^3$

$Volume = 366.3573$

$Volume = 366.36in^3$

<em>So, the edge length of the cube is 7.16 inches and the volume is 366.36 cubic inches</em>

<em />

6 0

3 years ago

Read 2 more answers

In Which set do all of the values make the inequality 2x-1<10 true

Step2247 [10]

X<5.5, x has to be less than 5.5 because you add 1 to both sides and then divided both sides by two and you would get x < 5.5

8 0

4 years ago

Simplify 9x^8 divided by 12x^5.

Dmitry_Shevchenko [17]

Ok, so rembmer some basica exponential rules and some fraction rules

(ab)/(cd)=(a/c)(b/d)
and
$\frac{x^n}{x^m} =x^{n_m}$

so
$\frac{9x^8}{12x^5}=( \frac{9}{12} )( \frac{x^8}{x^5} )=( \frac{3}{4} )( x^{8-5} )= ( \frac{3}{4} )( x^{3} )= \frac{3x^3}{4}$

6 0

3 years ago

The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e

irakobra [83]

Answer:

We need a sample size of at least 383.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

85% confidence level

So $\alpha = 0.15$ , z is the value of Z that has a pvalue of $1 - \frac{0.15}{2} = 0.925$ , so $Z = 1.44$ .

How large a sample would be required in order to estimate the fraction of tenth graders reading at or below the eighth grade level at the 85% confidence level with an error of at most 0.03

We need a sample size of at least n.

n is found with $M = 0.03, \pi = 0.21$

Then

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.03 = 1.44\sqrt{\frac{0.21*0.79}{n}}$

$0.03\sqrt{n} = 1.44\sqrt{0.21*0.79}$

$\sqrt{n} = \frac{1.44\sqrt{0.21*0.79}}{0.03}$

$(\sqrt{n})^{2} = (\frac{1.44\sqrt{0.21*0.79}}{0.03})^{2}$

$n = 382.23$

Rounding up

We need a sample size of at least 383.

6 0

3 years ago

Write an augmented matrix for the following system of equations

chubhunter [2.5K]

Just use the coefficients for each variable and put it in a matrix. Is the second question set equal to something? Because then you would put it in place as the question mark. (Did the pic come through?)
So the 3x3 matrix is made up of the coefficients and then to augment the matrix, you add another column at the end with the solutions. Good luck!

6 0

3 years ago