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3 years ago

Which of these relations are functions ?

Mathematics

2 answers:

Alex17521 [72]3 years ago

7 0

Your answer is 3. Y=2x-5

elixir [45]3 years ago

4 0

Answer:

A) y=5

C) y=2x-5

D) 2y=x-4

Step-by-step explanation:

You might be interested in

Find sin θ if θ is in Quadrant III and tan θ = . 0.958

Leokris [45]

Use the following identities:
$sec^2 = 1 + tan^2 \\ \\ sec = \frac{1}{cos} \\ \\ sin^2 = 1 - cos^2$
Also because the angle is in quadrant 3, sin must be negative.
Therefore
$sin = - \sqrt{1 - \frac{1}{1 + tan^2}}$
Subbing in tan = 0.958
$sin \theta = -0.69178$

7 0

3 years ago

Simplify 4/(√5+ √2) - 3/(√5- √2)

MakcuM [25]

Step-by-step explanation:

$\frac{4}{ \sqrt{5} + \sqrt{2}} - \frac{3}{ \sqrt{5} - \sqrt{2} } = \\ = \frac{4( \sqrt{5} - \sqrt{2} ) - 3( \sqrt{5} + \sqrt{2} ) }{( \sqrt{5} + \sqrt{2} )\times ( \sqrt{5} - \sqrt{2} )} = \\ = \frac{4 \sqrt{5} - 4 \sqrt{2} - 3 \sqrt{5} - 3 \sqrt{2} }{5 - \sqrt{10} + \sqrt{10} - 2 } = \\ = \frac{ \sqrt{5} - 7 \sqrt{2} }{3}$

8 0

3 years ago

A recent study from the University of Virginia looked at the effectiveness of an online sleep therapy program in treating insomn

Ludmilka [50]

Answer:

1. The 99% confidence interval for the difference in average is -6.47377 < μ₁ - μ₂ < -11.34623

2. The possible issues in the calculations includes;

a. The confidence level used in the confidence interval can influence the result of the confidence interval observed

b. The sample size is small

Step-by-step explanation:

1. The number of adults with insomnia in the sample = 45

The number of adults that participated in the therapy, n₁ = 22

The number of candidates that served as control group, n₂ = 23

The average score for the for the 22 participants of the program, $\overline x_1$ = 6.59

The standard deviation for the 22 participants of the program, s₁ = 4.10

The average score for the for the 23 subjects in the control group, $\overline x_2$ = 15.50

The standard deviation for the 23 subjects in the control group, s₂ = 5.34

The confidence interval for unknown standard deviation, σ, is given by the following expression;

$\left (\bar{x}_{1}- \bar{x}_{2} \right )\pm t_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}$

α = 1 - 0.99 = 0.01

α/2 = 0.005

The degrees of freedom, df = 22 - 1 = 21

$t_{\alpha /2}$ = $t_{0.005, \, 21}$ = 1.721

Therefore, we have;

$\left (6.59- 15.5 \right )\pm1.721 \cdot \sqrt{\dfrac{4.10^{2}}{22}+\dfrac{5.34^{2}}{23}}$

The 99% confidence interval for the difference in average is therefore given as follows;

-6.47377 < μ₁ - μ₂ < -11.34623

Therefore, there is considerable evidence that the participants in the survey had lower average score than the subjects in the control group

2. The possible issues in the calculations are;

a. The confidence level used in the confidence interval can influence the result of the confidence interval observed

b. The sample size is small

5 0

3 years ago

Ronin kicks a ball over the gate of the first floor balcony. The function h(t)=-36t^2+288t+9 can be used to model the height of

ale4655 [162]

Answer: 4 seconds

Step-by-step explanation:

Given: The function $h(t)=-36t^2+288t+9$ can be used to model the height of the ball where t is the time in seconds after the ball kicked and h(t) is the height in feet.