All categories

3 years ago

Is this graph a function?

Mathematics

1 answer:

Alex17521 [72]3 years ago

5 0

No.

To check if a graph is a function, make sure there is exactly one y value for every x value. To find this, pretend you are moving a vertical line across the entire graph. As long as it stays straight up and down, it should only touch the line in one place.

For example, put an imaginary vertical line at the x value of -4. There are 3 different places the lines touch, so this graph doesn’t represent a function.

You might be interested in

I know this isn't like a homework questions but what can help you to fall asleep? I have to get up hecka early and I can't go to

Elena-2011 [213]

Possibly Melatonin if you have any, Music can help sometimes relax your mind, reading can help tire you out, and/or Television can also relax your mind and tire you out. Hopefully this helps!

8 0

4 years ago

Read 2 more answers

Use a t-distribution to find a confidence interval for the difference in means μd=μ1-μ2 using the relevant sample results from p

V125BC [204]

Answer:

best estimate = Xd[bar]=4.80

margin of error = 8.66

The 99% confidence interval is -3.86 to 13.46

test statistic = -1.86

p-value = 0.0526

Decision: Reject the null hypothesis.

At the 5% significance level, you can conclude that the population mean of the difference between treatment 1 and treatment 2 is less than zero.

Step-by-step explanation:

Hello!

a) 99% CI

Using d=X₁-X₂ to determine the study variable Xd: the difference between treatment 1 and treatment 2.

Assuming that this variable has an approximately normal distribution: Xd≈N(μd;σ²d)

To calculate the sample mean and standard deviation you have to calculate the difference between the values of both treatments first.

Case 1 ; Case 2 ; Case 3 ; Case 4 ; Case 5

22-18= 4 ; 27-29= -2 ; 32-25= 7; 26-20= 6 ; 29-20= 9

n= 5

Xd[bar]= ∑X/n= 24/5= 4.80

Sd²= 1/(n-1)*[∑X²-(∑X)²/n]= 1/4*[186-(24²)/5]= 17.7

Sd= 4.21

The parameter of interes is the population mean od the difference, μd

The best estimate for this parameter is the sample mean, Xd[bar]=4.80

Using the t-distribution, the formula for the Confidence Interval is

Xd[bar] ± $t_{n-1;1-\alpha /2}$ * $\frac{Sd}{\sqrt{n} }$

Where the margin of error is:

$t_{n-1;1-\alpha /2}$ * $\frac{Sd}{\sqrt{n} }$ = $t_{4;0.995}$ * $\frac{Sd}{\sqrt{n} }$ = 4.604* $\frac{4.21}{\sqrt{5} }$ = 8.66

99% CI [-3.86; 13.46]

b) 5% Hypothesis test

The variable of interest is defined d=X₁-X₂; Xd: the difference between treatment 1 and treatment 2. Xd≈N(μd;σ²d)

The statistic hypotheses are:

H₀: μd = 0

H₁: μd < 0

α: 0.05

The statistic to use for this test is:

$t_{H_0}= \frac{X_d[bar]-Mu_d}{\frac{Sd}{\sqrt{n} } } ~~t_{n-1}$

As before you have to calculate the difference between the observation for each case and then the sample mean and standard deviation:

Case 1 ; Case 2 ; Case 3 ; Case 4 ; Case 5 ; Case 6 ; Case 7 ; Case 8

18-18= 0; 12-19= -7; 11-25= -14; 21-21= 0; 15-19= -4; 11-14=-3; 14-15= -1; 22-20= 2

n= 8

Xd[bar]= ∑X/n= -27/8= -3.38

Sd²= 1/(n-1)*[∑X²-(∑X)²/n]= 1/7*[275-(-27²)/8]= 26.27

Sd= 5.13

$t_{H_0}= \frac{-3.38-0}{\frac{5.13}{\sqrt{8} } }= -1.86$

This test is one-tailed to the left, which means that you will reject the null hypothesis to small values of t, the p-value of the test has the same direction as the rejection region, this means that it is one-tailed to the left and you can calculate it as:

P(≤-1.86)= 0.0526

The decision rule using the p-value is:

If p-value > α, do not reject the null hypothesis.

If p-value ≤ α, reject the null hypothesis.

The p-value is greater than the significance level so the decision is to reject the null hypothesis.

I hope it helps!