If we convert the given in its mathematical form, we have,
(30x⁶/14y⁵)(7y²/6x⁴)
It can be observed that the numerator of the first and the denominator of the second have a common factor of 6x⁴. Also, the denominator of the first and the numerator of the second expression have a common factor of 7y².
((6x⁴)(5x²)/(7y²)(2y³))(7y²/6x⁴)
Cancellation of the common terms will give us an answer of,
<em>5x²/2y³
</em><em />Therefore, the simplified version of the involved operation is 5x²/2y³. <em>
</em>
Answer:
6 is the domain because it is always the x value while range is the y value.
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.
The least common denominator is 1.
