for a graph to have three local maxima
if you notice that graphs maxima and minima are 1 less that its degree
x² has one turn or minima, it looks like ∪
x^ 3 has 2 turns
so to have three maxima its likely a x^4 or 4 degree equation
Answer:
16 acres
Step-by-step explanation:
First find the operating cost for 1 acre. Find that by dividing $551,520 by 20 acres. That gives us $27,576 per acre. Now we can set up a proportion to solve for the number of acres that have an operating cost of $441,216:
Cross multiply to get:
27576x = 441216 so
x = 16
We did not have to find the cost per acre to solve this. We could have set up the proportion with $551,520 on the top of the ratio and 20 on the bottom and we would still have gotten the same answer. I just figured the numbers wouldn't be as large if we divided to find the cost per unit.
Volume a sphere: [4/3]π(r^3)
Space between the spheres = Volume of the larger sphere - Volume of the smaller sphere
= [4/3]π (R^3) - [4/3π](r^3) = [4/3]π(R^3 - r^3) = [4/3]π {(5cm)^3 - (4cm)^3} =
= 255.5 cm^3
Answer: 255.5 cm^3
Answer:
thus the probability that a part was received from supplier Z , given that is defective is 5/6 (83.33%)
Step-by-step explanation:
denoting A= a piece is defective , Bi = a piece is defective from the i-th supplier and Ci= choosing a piece from the the i-th supplier
then
P(A)= ∑ P(Bi)*P(C) with i from 1 to 3
P(A)= ∑ 5/100 * 24/100 + 10/100 * 36/100 + 6/100 * 40/100 = 9/125
from the theorem of Bayes
P(Cz/A)= P(Cz∩A)/P(A)
where
P(Cz/A) = probability of choosing a piece from Z , given that a defective part was obtained
P(Cz∩A)= probability of choosing a piece from Z that is defective = P(Bz) = 6/100
therefore
P(Cz/A)= P(Cz∩A)/P(A) = P(Bz)/P(A)= 6/100/(9/125) = 5/6 (83.33%)
thus the probability that a part was received from supplier Z , given that is defective is 5/6 (83.33%)
To do this, you got to square 256.
The square root of 256 is 16.
Therefore, there are 16 small squares on each edge of the mosaic.
Kinda proof:
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25 squares. Square root is 5. 5 along each edge. My work shares same concept.
Extremely unnecessary proof:
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There are 256 squares, and you can count 16 on each edge. this shows 16 times 16, or 16 squared, which is 256.
