Answer:
Step-by-step explanation:
Corresponding scores before and after taking the course form matched pairs.
The data for the test are the differences between the scores before and after taking the course.
μd = scores before taking the course minus scores before taking the course.
a) For the null hypothesis
H0: μd ≥ 0
For the alternative hypothesis
H1: μd < 0
b) We would assume a significance level of 0.05. The P-value from the test is 0.65. The p value is high. It increases the possibility of accepting the null hypothesis.
Since alpha, 0.05 < than the p value, 0.65, then we would fail to reject the null hypothesis. Therefore, it does not provide enough evidence that scores after the course are greater than the scores before the course.
c) The mean difference for the sample scores is greater than or equal to zero
<span>3^5 * 6^-6 / 3^3 * 6^-4
= 3^2 6^-2
= 3^2 / 6^2
= 9/36
=1/4
answer
</span><span>C.
1/4</span>
To reduce a fraction, divide the numerator and the denominator equally until they reach the simplest whole number possible.
In this case, the numerator (720) and the denominator (1080) can both be divided by 360 to get 2/3, our reduced fraction.
Answer:A solution to an equation is the value or values of the variable or variables that make the equation a true statement. Graphically, solutions are the intersections of the graphs of the left side and the right side, or if the equation is written so that one side is zero, we are looking for the x-intercepts (for real solutions.)
Periodic functions can have infinite solutions. For instance, cos(x)=1 has as solutions x=2n*pi, n in ZZ (or n an integer.) Periodic functions can...
Step-by-step explanation:
