Answer:https://quiz.let.com/216018638/course-1-chapter-4-multiply-divide-fractions-6ns1-6rp3-3d-flash-cards/
remove the period between quiz and let but there is a quizlet on it
Step-by-step explanation:
If the ground beef is in pounds, it would be 5.25 lbs of ground beef.
We would get this answer by converting 3/4 into a decimal, .75. Then we multiply .75 by 7 (the number of people).
((This is if the amount of ground beef is in pounds, I'm not sure what it would be if it wasn't))
The answer is:
"<span>The degree of overlap between the two distributions is moderate."
I just took the test......</span>
Answer:
Hence, the limit of the expression:
is:
![\dfrac{1}{4}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B4%7D)
Step-by-step explanation:
We have to estimate the limit of:
![\lim_{x \to 3} \dfrac{x-3}{x^2-2x-3}](https://tex.z-dn.net/?f=%5Clim_%7Bx%20%5Cto%203%7D%20%5Cdfrac%7Bx-3%7D%7Bx%5E2-2x-3%7D)
We can also represent the denominator of the function in the limit as:
![x^2-2x-3=x^2-3x+x-3\\\\x^2-2x-3=x(x-3)+1(x-3)\\\\x^2-2x-3=(x+1)(x-3)](https://tex.z-dn.net/?f=x%5E2-2x-3%3Dx%5E2-3x%2Bx-3%5C%5C%5C%5Cx%5E2-2x-3%3Dx%28x-3%29%2B1%28x-3%29%5C%5C%5C%5Cx%5E2-2x-3%3D%28x%2B1%29%28x-3%29)
Hence, we have to estimate the limit of:
![\lim_{x \to 3} \dfrac{x-3}{(x+1)(x-3)}\\\\= \lim_{x \to 3}\dfrac{1}{x+1}\\ \\=\dfrac{1}{3+1}\\\\=\dfrac{1}{4}](https://tex.z-dn.net/?f=%5Clim_%7Bx%20%5Cto%203%7D%20%5Cdfrac%7Bx-3%7D%7B%28x%2B1%29%28x-3%29%7D%5C%5C%5C%5C%3D%20%5Clim_%7Bx%20%5Cto%203%7D%5Cdfrac%7B1%7D%7Bx%2B1%7D%5C%5C%20%5C%5C%3D%5Cdfrac%7B1%7D%7B3%2B1%7D%5C%5C%5C%5C%3D%5Cdfrac%7B1%7D%7B4%7D)
Hence, the limit of the expression:
is:
![\dfrac{1}{4}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B4%7D)
Answer:
- the way its set up its kind of confusing ↑
Step-by-step explanation: