Answer: partial correlation.
Explanation:
partial correlation measures the degree of association between two random variables, with the effect of a set of controlling random variables removed. If we are interested in finding to what extent there is a numerical relationship between two variables of interest, using their correlation coefficient will give misleading results if there is another, confounding, variable that is numerically related to both variables of interest. This misleading information can be avoided by controlling for the confounding variable, which is done by computing the partial correlation coefficient.
Answer:
x= 10
Step-by-step explanation:
Since the triangles are congruent
<B = <E
50 = 5x
Divide each side by 5
50/5 = 5x/5
10 =x
V = (4/3) pi r^3
9. V = (4/3)(3.14)(7.62)^3 = 1852.4 meters^3
10. V = (4/3)(3.14)(33/2)^3 = 18,807.0 inches^3
11. V = (4/3)(3.14)(18.4/2)^3 = 3260.1 feet^3
12. V = (4/3)(3.14)(sqrt3)/2)^3 = 2.7 cm^3
13. C = 2*pi*r ; 24 = 2 * 3.14 * r; 24/6.28 = r ; r = 3.82
V = (4/3)(3.14)(3.82)^3 = 233.4 in^3
14.V = (4/3)(3.14)(35.8)^3 = 192,095.6 mm^3
I can't read # 15 but follow the steps above.
Answer: 0.0548
Step-by-step explanation:
Given, A research study investigated differences between male and female students. Based on the study results, we can assume the population mean and standard deviation for the GPA of male students are µ = 3.5 and σ = 0.05.
Let 
represents the sample mean GPA for each student.
Then, the probability that the random sample of 100 male students has a mean GPA greater than 3.42:
![P(\overline{X}>3.42)=P(\dfrac{\overline{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}>\dfrac{3.42-3.5}{\dfrac{0.5}{\sqrt{100}}})\\\\=P(Z>\dfrac{-0.08}{\dfrac{0.5}{10}})\ \ \ [Z=\dfrac{\overline{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}]\\\\=P(Z>1.6)\\\\=1-P(Z](https://tex.z-dn.net/?f=P%28%5Coverline%7BX%7D%3E3.42%29%3DP%28%5Cdfrac%7B%5Coverline%7BX%7D-%5Cmu%7D%7B%5Cdfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D%3E%5Cdfrac%7B3.42-3.5%7D%7B%5Cdfrac%7B0.5%7D%7B%5Csqrt%7B100%7D%7D%7D%29%5C%5C%5C%5C%3DP%28Z%3E%5Cdfrac%7B-0.08%7D%7B%5Cdfrac%7B0.5%7D%7B10%7D%7D%29%5C%20%5C%20%5C%20%5BZ%3D%5Cdfrac%7B%5Coverline%7BX%7D-%5Cmu%7D%7B%5Cdfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D%5D%5C%5C%5C%5C%3DP%28Z%3E1.6%29%5C%5C%5C%5C%3D1-P%28Z%3C1.6%29%5C%5C%5C%5C%3D1-0.9452%3D0.0548)
hence, the required probability is 0.0548.
I cant see anything sorry
