The decision rule for rejecting the null hypothesis, considering the t-distribution, is of:
- |t| < 1.9801 -> do not reject the null hypothesis.
- |t| > 1.9801 -> reject the null hypothesis.
<h3>What are the hypothesis tested?</h3>
At the null hypothesis, it is tested if there is not enough evidence to conclude that the mean voltage for these two types of batteries is different, that is, the subtraction of the sample means is of zero, hence:
At the alternative hypothesis, it is tested if there is enough evidence to conclude that the mean voltage for these two types of batteries is different, that is, the subtraction of the sample means different of zero, hence:
We have a two-tailed test, as we are testing if the mean is different of a value.
Considering the significance level of 0.05, with 75 + 46 - 2 = 119 df, the critical value for the test is given as follows:
|t| = 1.9801.
Hence the decision rule is:
- |t| < 1.9801 -> do not reject the null hypothesis.
- |t| > 1.9801 -> reject the null hypothesis.
More can be learned about the t-distribution in the test of an hypothesis at brainly.com/question/13873630
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Answer:
This is a polynomial.
Step-by-step explanation:
A monomial is a constant, a variable or the product of constants and variables.
A binomial is the sum or difference of two monomials.
A trinomial is the sum or difference of three monomials.
Anything more than three terms makes it a polynomial.
This has 4 terms; thus it is a polynomial.
I think it is 715 mph but I may have done my math wrong....
Answer:
Originally there was 324 nuts in the bag.
Phillip took 108, Joy took 54, Brent took 81 and Preston took 10.
Step-by-step explanation:
Let's call the total amount of nuts N, and the number of nuts each child took by the initial of their name (Phillip will be P1 and Preston will be P2).
So we can write the following equations:
P1 = N/3
After removing N/3, the remaining nuts is (N - N/3):
J = (N - N/3)/4 = (2N/3)/4 = N/6
After removing N/6 from (N - N/3), the remaining nuts is (N - N/3 - N/6):
B = (N - N/3 - N/6)/2 = (N/2)/2 = N/4
P2 = 10
In the final there were 71 nuts remaining, so we have that:
N - N/3 - N/6 - N/4 - 10 = 71
N - N/3 - N/6 - N/4 = 81
N/4 = 81
N = 324 nuts
The amount of nuts took by each child is:
P1 = N/3 = 108 nuts
J = N/6 = 54 nuts
B = N/4 = 81 nuts
P2 = 10 nuts
