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Orlov [11]
2 years ago
6

Find the average rate of change over the interval [-1,2] (from x = -1 to x = 2) for the graph below

Mathematics
1 answer:
Lyrx [107]2 years ago
8 0

Hello!

Find the coordinate where x = -1:

At x = -1, y = 1, so the coordinate is (-1, 1).

Find the coordinate where x = 2:

At x = 2, y = -2, so the coordinate is (2, -2).

Find the rate of change between the points using the slope formula:

slope = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}

Plug in the coordinates above:

slope = \frac{-2-1}{2-(-1)}

Simplify:

slope = \frac{-3}{3} = -1

Thus, the rate of change between the points is -1.

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737 miles

Step-by-step explanation:

So we know that he drove 871 miles in 13 hours so first you want to do 871 divided by 13 and get 67 which means he drove 67 miles in one hour.  Finally you just have to multiply 67 by 11 and get 737.

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Consider the following data set. The data were actually collected in pairs, and each row represents a pair.
Anna11 [10]

Answer:

Step-by-step explanation:

Hello!

a.

Using the data sets you have to analyze them as if they are two independent samples.

The hypotheses are:

H₀: μ₁=μ₂

H₁: μ₁≠μ₂

α: 0.05

Assuming that both data sets are from a normal distribution and both population variances, although unknown, are equal, the statistic to use is:

t=\frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ~~t_{n_1+n_2-2}

Sa^2= \frac{(n_1-1)S^2_1+(n_2-1)S^2_2}{n_1+n_2-2}

Sa^2= \frac{9*5.38+9*3.65}{10+10-2} =4.51

Sa= 2.12

t_{H_0}= \frac{(49.69-50.52)-0}{2.12*\sqrt{\frac{1}{10} +\frac{1}{10} } } = -0.875

The p-value for the two-tailed test is:

P(t_{18}≤-0.875) + P(t_{18}≥0.875)= P(t_{18}≤-0.875) + (1 - P(t_{18}≤0.875))= 0.1965 + ( 1 - 0.8035)= 0.393

Since there is no significant level I determined it at 5%, comparing it to the p-value, the hypothesis test is not significant. Meaning that the difference between the population means of both groups is equal to zero.

b.

Considering the given data as a paired sample, you have to determine the variable difference to conduct the test:

Xd: the difference between X₁ and X₂

Before the sample mean and sample standard deviation you have to calculate the difference between the observations of group 1 and group 2. (2nd attachment)

Mean X[bar]d= -0.83

Standard deviation Sd= 1.28

The hypotheses for the paired sample test are:

H₀: μd=0

H₁: μd≠0

α: 0.05

t= \frac{X[bar]d-Mud}{\frac{Sd}{\sqrt{n} } } ~~t_{n-1}

t_{H_0}= \frac{-0.83-0}{\frac{1.28}{ \sqrt{10} }} = -2.05

The p-value for the two-tailed test is:

P(t_{9}≤-2.05) + P(t_{9}≥2.05)= P(t_{9}≤-2.05) + (1 - P(t_{9}≤2.05))= 0.0353 + (1 - 0.9647)= 0.0706

Comparing the p-value with the significance level, the hypothesis test is not significant. Meaning that the population mean of the difference between group 1 and group 2 is equal to cero. There is no difference between the two groups.

c.

The value of the statistic in "a" is greater than the value of the statistic obtained in "b". Since there are two samples used an "a" the degrees of freedom of the test are the double as the ones used in "b". The p-value obtained in "b" is around half of the p-value of "a".

The main difference is that in "a" you compared two different samples but in "b" you compared two dependent samples when analyzing paired data the "effect of the individual" is removed from the equation.

I hope this helps!

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Answer = 5(x+1)^2 -21

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5 {(x+1)^2 -1 } - 16
5(x+1)^2 -5 - 16

5(x+1)^2 -21
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