The denominator of the second fraction can be factored as (a-2)(2a-7), then it becomes doable. You cancel the (2a-7) factors, and are left with:
Do note that you have erased the fact that a≠0 and a≠7/2, so you should always mention that.
Answer:
the second one is the right answer (2)
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.
If it is 5ft a minute then it must be 10ft every two minutes
Step-by-step explanation:
30÷6 =5 or 30÷5 =6 good luck!
