when solving |5k+6|=39 what two equations would we set up to allow us to get rid of the absolute value bars?
1 answer:
You might be interested in
Answer:
(2, 7)
Step-by-step explanation:
represents the reflection of a point P about the y-axis.
If a point (x, y) is reflected across y-axis, rule for the reflection is,
(x, y) → (-x, y)
If a point (-2, 7) follows the same rule, coordinates of the image point will be,
P(-2, 7) → P'(2, 7)
Therefore, answer is (2, 7).
Answer:
The probability of obtaining a result equivalent to or greater than what was the truly observed value of the test statistic is 0.001.
Step-by-step explanation:
The researcher can use a one sample <em>z</em> test to determine whether his claim is correct or not.
The hypothesis to test whether the mean BMI is more than 25.0, is defined as:
<em>H₀</em>: The mean BMI is not more than 25.0, i.e. <em>μ </em>≤ 25.0.
<em>Hₐ</em>: The mean BMI is more than 25.0, i.e. <em>μ </em>> 25.0.
The test statistic is defined as:
The <em>p</em>-value of the test is,
<em>p</em> = 0.001.
The p-value is well-defined as per the probability, [under the null-hypothesis (<em>H₀</em>)], of attaining a result equivalent to or greater than what was the truly observed value of the test statistic.
A small p-value (typically ≤ 0.05) specifies sufficient proof against the null hypothesis (<em>H₀</em>), so you discard <em>H₀</em>. A large p-value (> 0.05) specifies fragile proof against the <em>H₀</em>, so you fail to discard <em>H₀.</em>
The <em>p</em>-value of 0.001 indicates that the probability of obtaining a result equivalent to or greater than what was the truly observed value of the test statistic is 0.001.
As the <em>p</em>-value is very small, the null hypothesis will be rejected at any level of significance.
Thus, concluding that the mean BMI of adult Canadians is more than 25.0.