Answer:
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Answer:
Step-by-step explanation:
This linear graph will be shifted downward by 2 1/2 units: first, from y = 1/2 to the x-axis, and next, from the x-axis by 2 more units, for a total shift of 2 1/2 units.
Answer:
a^2 + b^2 = c^2
Step-by-step explanation:
a^2 + b^2 = c^2
Answer:
A. 1/5k - 2/3j and -2/3j +1/5k
Step-by-step explanation:
A. 1/5k - 2/3j and -2/3j +1/5k
B. 1/5k - 2/3j and -1/5k +2/3j
There is a change in the signs of each term
1/5k changed to -1/5k
-2/3j changed to +2/3j
Not equivalent
C. 1/5k - 2/3j and 1/5j - 2/3k
There is a change in the variables
1/5k changed to 1/5j
-2/3j changed to -2/3k
D. 1/5k - 2/3j and 2/3j - 1/5k
The is a change in the signs of each term
1/5k changed to -1/5k
-2/3j changed to +2/3j
The only equivalent expression is
A. 1/5k - 2/3j and -2/3j +1/5k
Answer:
e. The probability of observing a sample mean of 5.11 or less, or of 5.29 or more, is 0.018 if the true mean is 5.2.
Step-by-step explanation:
We have a two-tailed one sample t-test.
The null hypothesis claims that the pH is not significantly different from 5.2.
The alternative hypothesis is that the mean pH is significantly different from 5.2.
The sample mean pH is 5.11, with a sample size of n=50.
The P-value of the test is 0.018.
This P-value corresponds to the probability of observing a sample mean of 5.11 or less, given that the population is defined by the null hypothesis (mean=5.2).
As this test is two-tailed, it also includes the probability of the other tail. That is the probability of observing a sample with mean 5.29 or more (0.09 or more from the population mean).
Then, we can say that, if the true mean is 5.2, there is a probability P=0.018 of observing a sample of size n=50 with a sample mean with a difference bigger than 0.09 from the population mean of the null hypothesis (5.11 or less or 5.29 or more).
The right answer is e.
