Answer:
Step-by-step explanation:
Let
x------> the length side of the square base of the box
y-------> the height of the box
we know that
volume of the box=b²*h
b=x
h=y
volume=256 cm³
so
256=x²*y------>y=256/x²--------> equation 1
<span>The amount of material used is directly proportional to the surface area, so we will minimize the amount of material by minimizing the surface area.</span>
surface area of the cardboard=area of the base+perimeter of base*height
area of the base=x²
perimeter of the base=4*x
height=y
surface area=x²+4x*y-----> equation 2
substitute equation 1 in equation 2
SA=x²+4x*[256/x²]-----> SA=x²+1024/x
step 1
find the first derivative of SA and equate to zero
2x+1024*(-1)/x²=0------> 2x=1024/x²----> x³=512--------> x=8 cm
y=256/x²------> y=256/8²-----> y=4 cm
the answer is
the length side of the square base of the box is 8 cm
the height of the box is 4 cm
Answer:
10 oranges
Step-by-step explanation:
4 oranges are needed to make one cup. 2 cups is 8 oranges and 1/2 cup is 2 oranges
Answer:
0.8 cm
Step-by-step explanation:
The formula for MAD is attached in the picture.
x_i are the values (data)
x bar is the mean
n is the number of numbers (data)
For figuring out MAD, basically, <em>we subtract the mean from each of the values respectively and take the absolute value and SUM for all of the numbers in the data set. Then we divide by n, the number of numbers.</em>
<em />
Let's do this:
MAD = 
MAD is 0.8
Answer:
a) 30 kangaroos in 2030
b) decreasing 8% per year
c) large t results in fractional kangaroos: P(100) ≈ 1/55 kangaroo
Step-by-step explanation:
We assume your equation is supposed to be ...
P(t) = 76(0.92^t)
__
a) P(10) = 76(0.92^10) = 76(0.4344) = 30.01 ≈ 30
In the year 2030, the population of kangaroos in the province is modeled to be 30.
__
b) The population is decreasing. The base 0.92 of the exponent t is the cause. The population is changing by 0.92 -1 = -0.08 = -8% each year.
The population is decreasing by 8% each year.
__
c) The model loses its value once the population drops below 1/2 kangaroo. For large values of t, it predicts only fractional kangaroos, hence is not realistic.
P(100) = 75(0.92^100) = 76(0.0002392)
P(100) ≈ 0.0182, about 1/55th of a kangaroo
