Answer:
The probability of selecting a sample which doesn't capture the true value of μ would be 10% rather than 5% if they decide to calculate a 90% confidence interval rather than a 95% confidence interval from the sample they will select.
Step-by-step explanation:
The correct answer is the last statement, but first, let's look at the other statements:
<em>Their confidence interval would be less likely to capture the </em><em>sample </em><em>mean.</em>
This statement is not correct because the <em>confidence </em>of a confidence interval gives us the probability of capturing the true value of the population mean.
<em>They would </em><em>decrease </em><em>the margin of error of their confidence interval if they calculated a 90% rather than a 95% confidence interval.</em>
If we want more confidence we must establish more precision, which means more error. In other words if the confidence increases so does the error for a fixed sample size. The second statement is false.
<em>The probability of selecting a sample which doesn't capture the true value of μ would be 10% rather than 5% if they decide to calculate a 90% confidence interval rather than a 95% confidence interval from the sample they will select.</em>
As stated before the <em>confidence </em>of a confidence interval is the probability that the interval contains the true value of μ. Therefore, if we increase the confidence from 95% to 90% then this probability also increases. A 90% confidence means that there's a 10% probability of not containing the true mean. Likely, a 95% confidence means that there's a 5% probability of not containig the true mean. The third statement is true.
Answer: 12
Step-by-step explanation:
2y - x = 5
x^2 + y^2 - 25 = 0
x = 2y - 5
(2y-5)^2 + y^2 - 25 = 0
(2y-5)(2y-5) + y^2 - 25 = 0
4y^2 - 20y + 25 + y^2 - 25 = 0
5y^2 - 20y = 0
y = 0 , y = 4
x = 2y - 5 , when y = 0
x = - 5
x = 2y - 5 , when y = 4
x = 8 - 5
x = 3
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