<h3>
Answer:</h3>
- C. (9x -1)(x +4) = 9x² +35x -4
- B. 480
- A. P(t) = 4(1.019)^t
Step-by-step explanation:
1. See the attachment for the filled-in diagram. Adding the contents of the figure gives the sum at the bottom, matching selection C.
2. If we let "d" represent the length of the second volyage, then the total length of the two voyages is ...
... (d+43) + d = 1003
... 2d = 960 . . . . . . . subtract 43
... d = 480 . . . . . . . . divide by 2
The second voyage lasted 480 days.
3. 1.9% - 1.9/100 = 0.019. Adding this fraction to the original means the original is multiplied by 1 +0.019 = 1.019. Doing this multiplication each year for t years means the multiplier is (1.019)^t.
Since the starting value (in 1975) is 4 (billion), the population t years after that is ...
... P(t) = 4(1.019)^t
Answer:
The answer is below
Step-by-step explanation:
The volume of the cylinder with radius (r) and height (h) is given as:
Volume = πr²h
17 = πr²h
h = 17 / πr²
Let k represent the cost of the side and 2k the cost of the bottom.
The area of the side = 2πrh, hence the cost of the side = k(2πrh)
The area of the top and bottom = 2πr², hence the cost of the top and bottom = 2k(2πr²)
The total cost (T) = cost of side + cost of the top and bottom
T = k(2πrh) + 2k(2πr²)
T = 2kπ(rh + 2r²)
Put h = 17/πr²
T = 2kπ (r×17/πr² + 2r²)
T = 2kπ(17/πr + 2r²)
The minimal cost is at dT/dr = 0
![\frac{dT}{dr} =0=\frac{d}{dr} [2k\pi(\frac{17}{\pi r} +2r^2)]\\\\4r-\frac{17}{\pi r^2} =0\\\\4r=\frac{17}{\pi r^2}\\\\r^3=\frac{17}{4\pi }=1.35\\\\r=\sqrt[3]{1.35}\\\\r=1.1\ m\\\\h=\frac{17}{\pi r^2} =\frac{17}{\pi (1.1)^2} \\\\h=4.42\ m](https://tex.z-dn.net/?f=%5Cfrac%7BdT%7D%7Bdr%7D%20%3D0%3D%5Cfrac%7Bd%7D%7Bdr%7D%20%5B2k%5Cpi%28%5Cfrac%7B17%7D%7B%5Cpi%20r%7D%20%2B2r%5E2%29%5D%5C%5C%5C%5C4r-%5Cfrac%7B17%7D%7B%5Cpi%20r%5E2%7D%20%3D0%5C%5C%5C%5C4r%3D%5Cfrac%7B17%7D%7B%5Cpi%20r%5E2%7D%5C%5C%5C%5Cr%5E3%3D%5Cfrac%7B17%7D%7B4%5Cpi%20%7D%3D1.35%5C%5C%5C%5Cr%3D%5Csqrt%5B3%5D%7B1.35%7D%5C%5C%5C%5Cr%3D1.1%5C%20m%5C%5C%5C%5Ch%3D%5Cfrac%7B17%7D%7B%5Cpi%20r%5E2%7D%20%20%3D%5Cfrac%7B17%7D%7B%5Cpi%20%281.1%29%5E2%7D%20%5C%5C%5C%5Ch%3D4.42%5C%20m)
Answer:
14.89% or 0.1489
Step-by-step explanation:
First, find the z-score for both 70 and 80 minutes and their corresponding percentile.
For X = 70 minutes:
This z-score is equivalent to the 7.445 th percentile, so the probability of a student finishing this exam in less than 70 minutes is 7.445%
or X = 80 minutes:
This z-score is equivalent to the 92.555 th percentile, so the probability of a student finishing this exam in more than 80 minutes is 100- 92.555 = 7.445%
Therefore, the probability (P) of a student finishing this exam in less than 70 minutes or more than 80 minutes is:
P = 7.445% + 7.445% = 14.89%
Answer:
me dont know
Step-by-step explanation:
We assume you intend that cars shipped to Japan are modeled by
j(n) = 2·3^(n-1)
and that those shipped to Vietnam are modeled by
v(n) = 5 + 11·(n-1)
The value of j(n) will exceed the value of v(n) in
month 4.
