I think you meant to say
(as opposed to <em>x</em> approaching 2)
Since both the numerator and denominator are continuous at <em>t</em> = 2, the limit of the ratio is equal to a ratio of limits. In other words, the limit operator distributes over the quotient:
Because these expressions are continuous at <em>t</em> = 2, we can compute the limits by evaluating the limands directly at 2:
Answer:
And we can find the limits in order to consider values as significantly low and high like this:
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
Solution to the problem
For this case we can consider a value to be significantly low if we have that the z score is lower or equal to - 2 and we can consider a value to be significantly high if its z score is higher tor equal to 2.
For this case we have the mean and the deviation given:
And we can find the limits in order to consider values as significantly low and high like this:
Answer:
yes
Step-by-step explanation:
1.2.3.4