Area of rectangle + Area of semi circle
12 * 4 + (22/7 * 4^2)/2
48 + 25.14
73.14 yard square
You can either make a table using any numbers you would like (I would suggest -5 to 5) and then graphing the rule
ex. (7/2)(-5)-2 = -19.5 (I multiplied (7/2) by -5 and then subtracted 2)
Or you can put the rule in a graphing calculator and check the points from there
Answer: The volume of largest rectangular box is 4.5 units.
Step-by-step explanation:
Since we have given that
Volume = 
with subject to 
So, let 
So, Volume becomes,
Partially derivative wrt x and y we get that
By solving these two equations, we get that
So, 
So, Volume of largest rectangular box would be
Hence, the volume of largest rectangular box is 4.5 units.
Answer:
a) 615
b) 715
c) 344
Step-by-step explanation:
According to the Question,
- Given that, A study conducted by the Center for Population Economics at the University of Chicago studied the birth weights of 732 babies born in New York. The mean weight was 3311 grams with a standard deviation of 860 grams
- Since the distribution is approximately bell-shaped, we can use the normal distribution and calculate the Z scores for each scenario.
Z = (x - mean)/standard deviation
Now,
For x = 4171, Z = (4171 - 3311)/860 = 1
- P(Z < 1) using Z table for areas for the standard normal distribution, you will get 0.8413.
Next, multiply that by the sample size of 732.
- Therefore 732(0.8413) = 615.8316, so approximately 615 will weigh less than 4171
- For part b, use the same method except x is now 1591.
Z = (1581 - 3311)/860 = -2
- P(Z > -2) , using the Z table is 1 - 0.0228 = 0.9772 . Now 732(0.9772) = 715.3104, so approximately 715 will weigh more than 1591.
- For part c, we now need to get two Z scores, one for 3311 and another for 5031.
Z1 = (3311 - 3311)/860 = 0
Z2 = (5031 - 3311)/860= 2
P(0 ≤ Z ≤ 2) = 0.9772 - 0.5000 = 0.4772
approximately 47% fall between 0 and 1 standard deviation, so take 0.47 times 732 ⇒ 732×0.47 = 344.
