Need help with geometry!! Thank you :)

PLEASE HELP ASAP I WILL GIVE BRAINSLET TO THE CORRECT ANSWER!!!!!!!!!!

cestrela7 [59]

<u>Answer</u>:

Therefore has one solution.

<u>explanation</u>:

Given equations:

6x + y = -7

-24x -7y = 25

Make y the subject:

6x + y = -7

y = -7 - 6x ..............equation 1

-24x -7y = 25

-7y = 25 + 24x

y = (25 + 24x)/-7 ..........equation 2

Solve them simultaneously:

(25 + 24x)/-7 = -7 - 6x

25 + 24x = -7 (-7 - 6x)

25 + 24x = 49 + 42x

42x - 24x = 25- 42

18x = -24

x = $-\frac{4}{3}$

Then y is:

y = -7 - 6x

y = -7 - 6( $-\frac{4}{3}$ )

y = 1

Has one separate value of x and y: ( $-\frac{4}{3}$ , 1 ); so it has one solution.

3 0

3 years ago

Of 150 adults selected randomly from one town, 30 of them smoke. Construct a 99% confidence interval for the true percentage of

anyanavicka [17]

Answer:

The 99% confidence interval for the true percentage of all adults in the town that smoke is (0.1159, 0.2841).

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 interval $1-\alpha$ , 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 $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

Of 150 adults selected randomly from one town, 30 of them smoke. This means that $n = 150, p = \frac{30}{150} = 0.2$

99% confidence interval

So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.20 - 2.575\sqrt{\frac{0.20*0.80}{150}} = 0.1159$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.20 + 2.575\sqrt{\frac{0.20*0.80}{150}}{119}} = 0.2841$

The 99% confidence interval for the true percentage of all adults in the town that smoke is (0.1159, 0.2841).

3 0

4 years ago

Solve this equation pls

Aleksandr-060686 [28]

Answer:

${ \rm{y = \frac{3 \sqrt{x} + 2x}{ {4x}^{2} } }} \\$

• We are gonna use the quotient rule

${ \boxed{ \bf{ \frac{dy}{dx} = \frac{ {\huge{v} } \frac{du}{dx} - { \huge{u}} \frac{dv}{dx} }{ {v}^{2} } }}} \\$

u is (3√x + 2x)
v is 4x²
du/dx is 3/2√x
dv/dx is 8x

• Therefore:

${ \rm{ \frac{dy}{dx} = \frac{(4 {x}^{2})( \frac{3}{2 \sqrt{x} }) - (3 \sqrt{x} + 2x)(8x) }{16 {x}^{4} } }} \\ \\ { \rm{ \frac{dy}{dx} = \frac{(6 {x}^{2})( {x}^{ - \frac{1}{2} } ) - (24 {x}^{ \frac{3}{2} } + 16 {x}^{2} ) }{16 {x}^{4} } }} \\ \\ { \rm{ \frac{dy}{dx} = \frac{ 6 {x}^{ \frac{3}{2} } - {24x}^{ \frac{3}{2} } - {16x}^{2} }{16 {x}^{4} } }} \\ \\ { \rm{ \frac{dy}{dx} = \frac{ - 18 {x}^{ \frac{3}{2} } - 16 {x}^{2} }{16 {x}^{4} } }} \\ \\ { \rm{ \frac{dy}{dx} = - 2 {x}^{ \frac{3}{2} } (9 + 8 {x}^{ \frac{4}{3} } ) \div 16 {x}^{4} }} \\ \\ { \boxed{ \boxed{ \rm{ \: \: \frac{dy}{dx} = - \frac{8 \sqrt{ {x}^{ \frac{8}{3} }} + 9}{8x {}^{ \frac{8}{3} } } \: \: }}}}$