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3 years ago

✨PLEASE HELP✨ DUE SOON

Mathematics

2 answers:

jasenka [17]3 years ago

7 0

Answer:

The answer is C!!

Step-by-step explanation:

4(2y-8)+y=13

8y-32+y=13

9y-32=13

9y=45

y=5

4x+y=13

4x+(5)=13

4x=8

x=2

mestny [16]3 years ago

6 0

Hey :) it’s (2,5) it’s C !!

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Lynna [10]

You have a function $y= \dfrac{3}{2} \left(1+\sin \left(\dfrac{2t+1}{2}\cdot \pi\right) \right)$ .
Since the range of the function $y=\sin x$ is $[-1,1]$ you have that

$-1\le \sin \left(\dfrac{2t+1}{2}\cdot \pi\right)\le 1 \\ 0\le 1+\sin \left(\dfrac{2t+1}{2}\cdot \pi\right) \le 2 \\ 0\le \dfrac{3}{2} \left(1+\sin \left(\dfrac{2t+1}{2}\cdot \pi\right) \right)\le 3$ .
Answer: The range of given function is [0,3] and the correct choice is C.

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3 years ago

Which of the following best describes the equation below? y=2/5x-12 A. neither linear nor nonlinear B. linear C. both linear and

patriot [66]

Answer:

B. Linear

Step-by-step explanation:

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3 years ago

If the ratio of the radii of two spheres is 2:3, the ratio of their surface areas will be 8:27.

Slav-nsk [51]

<span>The ratio of 2 similar areas is the square of the ratio of their RADIUS or (sides)
R²=(2/3)² = 4/9 This should be the answer

Whereas R³ = (2/3)³ = 8/27 is the ratio of their volume & not area

</span>

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3 years ago

Identify the percent of change as increase or decrease of 138 to 75

RUDIKE [14]

138 x ? = 75
? = 75/138 = 0.544(3sf)
so about 54.35% (2dp) decrease

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3 years ago

In a study of computer use, 1000 randomly selected Canadian Internet users were asked how much time they spend using the Interne

Dvinal [7]

Answer:

b. Do not reject H0. We do not have convincing evidence that the mean weekly time spent using the Internet by Canadians is greater than 12.7 hours.

Step-by-step explanation:

Given that in a study of computer use, 1000 randomly selected Canadian Internet users were asked how much time they spend using the Internet in a typical week. The mean of the sample observations was 12.9 hours.

$H_0: \bar x =12.7\\H_a: \bar x >12,7$

(Right tailed test at 5% level)

Mean difference = 0.2

Std error = $\frac{6}{\sqrt{1000} } \\=0.1897$

Z statistic = 1.0540

p value = 0.145941

since p >alpha we do not reject H0.

b. Do not reject H0. We do not have convincing evidence that the mean weekly time spent using the Internet by Canadians is greater than 12.7 hours.

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3 years ago