The second answer that uses = is not correct, because the pentagons are similar not congruent.
The third answer is not correct, because that sign means estimated, not similar.
I've never seen the fourth sign, so by process of elimination I believe the answer is the first one.
<span>ABCDE~QRSTU</span>
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.
Answer:
-6-(-1)
Step-by-step explanation:
Since all the boxes are negative, and there are 6 boxes, the first part of the equation is -6. Then since the last box is being taken away, we are subtracting it. The last box is still negative though so the last part is -1. Since -6 subtracted by -1 is -6-(-1), the answer is:
-6-(-1)
Step-by-step explanation:
Fractions with only factors of 2 and 5 are terminating. If not, they give repeating decimals.
Therefore 17/8 and 34/16 are terminating whereas 2/13, 5/24 and 6/7 are repeating decimals.
