The coordinates of the vertices of a rectangle are (-2, 3), (4, -4), (4, 3), and (-2, -4). What are the dimensions of the rectan
gle? A. 1 unit by 2 units B. 1 unit by 6 units C. 7 units by 2 units D. 7 units by 6 units
1 answer:
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Answer:
− 3 y ' ' − 3 y ' + 3 y = 0 : over-damped
− 2 y ' ' − 4 y ' + 1 y = 0 : over-damped
1 y ' ' + 7 y ' + 5 y = 0: over-damped
Step-by-step explanation:
Using the characteristic equation you can express a differential equation of order n as an algebraic equation of degree n:
This differential equation will have a characteristic equation of the form:
Now, you can classify the solution for a differential equation using a simple method. In order to do it, you just need to use the discriminant.
- If the discriminant is greater than zero, the solution is over-damped
- If the discriminant is less than zero, the solution is under-damped
- If the discriminant is equal to zero, the solution is critically damped
So, given the differential equation:
Which has characteristic equation of the form:
The quadratic polynomial of the form:
Has discriminant:
In this case:
So:
In this case:
Therefore the solution is over-damped.
Now, given the differential equation:
Which has characteristic equation of the form:
The quadratic polynomial of the form:
Has discriminant:
In this case:
So:
In this case:
Therefore the solution is over-damped.
Finally, given the differential equation:
Which has characteristic equation of the form:
The quadratic polynomial of the form:
Has discriminant:
In this case:
So:
In this case:
Therefore the solution is over-damped.
Regarding the hypothesis tested in this problem, it is found that:
a) The test statistic is: t = 0.49.
b) The p-value is: 0.6297..
c) The null hypothesis is not rejected, as the p-value is greater than the significance level.
d) It appears that the students are legitimately good at estimating one minute, as the sample does not give enough evidence to reject the null hypothesis.
<h3>What is the test statistic?</h3>The test statistic of the t-distribution is given by the equation presented as follows:
In which the variables of the equation are given as follows:
is the sample mean.
is the value tested at the hypotheses.
- s is the standard deviation of the sample.
- n is the sample size.
The value tested at the hypothesis is:
From the sample, using a calculator, the other parameters are given as follows:
Hence the test statistic is obtained as follows:
t = 0.49.
<h3>What are the p-value and the conclusion?</h3>The p-value is obtained using a t-distribution calculator, with a two-tailed test, as we are testing if the mean is different a value, in this case 60, with:
- 15 - 1 = 14 df, as the number of degrees of freedom is one less than the sample size.
- t = 0.49, as obtained above.
Hence the p-value is of 0.6297.
Since the p-value is greater than the significance level of 0.01, the null hypothesis is not rejected, attesting to the ability of the students to estimate one minute.
More can be learned about the test of an hypothesis at brainly.com/question/13873630
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