What is - 3x - ( - 2 - 4x ) if x equals 5

Can you please answer these 3 questions im having a hard time rn and i just need to relax but i really need to get this done, pl

Yanka [14]

Answer:

1. Infinitiely many solutions, one solution, no solutions

2.

y=-3/2x-4

y= x+1

3.

y=1/4x-2

y=1/4x+3

Step-by-step explanation:

4 0

3 years ago

Can someone else me with this

kolezko [41]

Answer:

8

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Janet has $75 to spend at the mall. She has already spent $28. She wants to buy games that cost $15 each, including tax. Enter t

serg [7]

Answer:

3 games

Step-by-step explanation:

The computation of the maximum number of games that he could buy is shown below:

= Amount available - already spent

= $75 - $28

= $47

If each game cost $15

So, the maximum no of games is

= $47 ÷ $15

= 3.133

= 3 games

3 0

3 years ago

Reading proficiency: An educator wants to construct a 99.5% confidence interval for the proportion of elementary school children

Shkiper50 [21]

Answer:

A sample size of 345 is needed so that the confidence interval will have a margin of error of 0.07

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 of the interval is given by:

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

In this problem, we have that:

$p = 0.69$

99.5% confidence level

So $\alpha = 0.005$ , z is the value of Z that has a pvalue of $1 - \frac{0.005}{2} = 0.9975$ , so $Z = 2.81$ .

Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.07?

This is n when M = 0.07. So

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

$0.07 = 2.81\sqrt{\frac{0.69*0.31}{n}}$

$0.07\sqrt{n} = 1.2996$

$\sqrt{n} = \frac{1.2996}{0.07}$

$\sqrt{n} = 18.5658$

$(\sqrt{n})^{2} = (18.5658)^{2}$

$n = 345$

A sample size of 345 is needed so that the confidence interval will have a margin of error of 0.07

4 0

3 years ago

99 POINT QUESTION, PLUS BRAINLIEST!!!

Basile [38]

1.Disc method.
In this method the volume is given by:

$\boxed{V=\pi\int\limits_a^b\big[f(x)\big]^2}$

so:

$V=\pi\int\limits_1^3x^4\,dx=\boxed{\pi\int\limits_1^3\big[x^2\big]^2\,dx}$

A) Function $f(x)=x^2$ over the interval $[1,3]$
B) We use disk method and f(x) is function of variable x, so we rotate the curve about the x-axis.

2. Shell method.

In this case volume is given by:

 $\boxed{V=2\pi\int\limits_a^bx\cdot f(x)\,dx}$

So there will be:

$V=\pi\int\limits_1^3x^4\,dx=\dfrac{2}{2}\cdot\pi\int\limits_1^3x^4\,dx=2\pi\int\limits_1^3\dfrac{x^4}{2}\,dx=
\boxed{2\pi\int\limits_1^3x\cdot\dfrac{x^3}{2}\,dx}$

A) Function $f(x)=\dfrac{x^3}{2}$ over the interval $[1,3]$
B) We use shell method and f(x) is function of variable x, so we rotate the curve about the y-axis.

3 0

3 years ago