Answer:
um i cannot understand that ok ?
Step-by-step explanation:
Answer:
Janice converted a mixed number to the wrong improper fraction.
Step-by-step explanation:
1 batch of stew uses 1 and 2/3 lb meat
Total amount of meat used - 10 and 5/6 lb
x = number of stew batches made
To get number of batches made, divide total meat used by the amount of meat each batch requires:
x = 10 5/6 ÷ 1 2/3
convert mixed fractions to improper fractions:
x = (10x6+5)/6 ÷ (1x3+2)/3
x = 65/6 ÷ 5/3
The fraction in bold numbers is where Janice got it wrong. She changed 1 2/3 to 5/2. (From the question: 10 and StartFraction 5 over 6 EndFraction divided by 1 and two-thirds = StartFraction 65 over 6 EndFraction divided by five halves). 1 2/3 converted to an improper fraction is 5/3. From this point going forward, the calculations are wrong.
Answer:
Sixty-three percent is the statistic and 39% is the parameter.
Step-by-step explanation:
Answer:
D.
Step-by-step explanation:
Remember that the limit definition of a derivative at a point is:
![\displaystyle{\frac{d}{dx}[f(a)]= \lim_{x \to a}\frac{f(x)-f(a)}{x-a}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%7B%5Cfrac%7Bd%7D%7Bdx%7D%5Bf%28a%29%5D%3D%20%5Clim_%7Bx%20%5Cto%20a%7D%5Cfrac%7Bf%28x%29-f%28a%29%7D%7Bx-a%7D%7D)
Hence, if we let f(x) be ln(x+1) and a be 1, this will yield:
![\displaystyle{\frac{d}{dx}[f(1)]= \lim_{x \to 1}\frac{\ln(x+1)-\ln(2)}{x-1}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%7B%5Cfrac%7Bd%7D%7Bdx%7D%5Bf%281%29%5D%3D%20%5Clim_%7Bx%20%5Cto%201%7D%5Cfrac%7B%5Cln%28x%2B1%29-%5Cln%282%29%7D%7Bx-1%7D%7D)
Hence, the limit is equivalent to the derivative of f(x) at x=1, or f’(1).
The answer will thus be D.
Answer:
Step-by-step explanation:
i) 1
ii) see attached
iii) as x goes from 0 to 2, y decreases parabolically from 1 to an acme at -3
iv) y = (x - 2)² - 3
v) -3 < t < 1
if t < -3 no intersection occurs between y = t and y = (x - 2)² - 3
if t = -3, y = t intersects the curve at only one point, (2, -3)
if -3 < t < 1 there exist two intersecting points with positive x values
if t = 1 there exist only one positive x value intersection as the other is at (0, 1) and 0 is neither positive nor negative.
if t > 1 there is one positive x value intersection and one negative x value intersection.