Question is Incomplete, Complete question is given below.
Mrs.Smith teaches junior high school. During a class party in fourth period, she used 4/5 liters of soda for 12 students. If she had a class party in fifth period and used the same serving size, how many liters did she use for 30 students?
Answer:
Mrs. Smith used 2 liters of soda for 30 students.
Step-by-step explanation:
Given:
Number of students = 12
Amount of soda used for 12 students = 4/5 liters
We need to find amount of soda used for 30 students.
For 12 students = 4/5 liters of soda used
For 30 students = the amount of soda used
By Using Unitary Method we get
Amount of soda used = 
Hence Mrs. Smith used 2 liters of soda for 30 students.
Answer:
The density of the pine log is 0.2842 pounds per inch.
Step-by-step explanation:
To calculate the density of an object, you must divide its mass by its volume (radius times height). Thus, given that the pine log that has a 5-inch radius and is 30 inches long, while its weight is approximately 42.63 pounds, the following calculation must be performed to determine its density:
X = 42.63 / (30 x 5)
X = 42.63 / 150
X = 0.2842
Thus, the density of the pine log is 0.2842 pounds per inch.
There are 4! = 24 poosible permutations of the four letters. Let the letters be A, B, C and D. Two permutations will have only letter A in the correct envelope, two more permutations will have only letter B in the correct envelope, two more will have only letter C in the correct envelope and two more will have only letter D in the correct envelope. Therefore 8 out of the 24 possible permutations will have only one letter with the correct address. The required probability is 8/24 = 1/3.
The first thing we are going to do for this case is define variables.
We have then:
y = the cost of the box
x = one side of the square base
z = height of the box
The volume of the building is 14,000 cubic feet:
x ^ 2 * z = 14000
We cleared z:
z = (14000 / x ^ 2)
On the other hand, the cost will be:
floor = 4 (x ^ 2)
roof = 3 (x ^ 2)
for the walls:
1 side = 16 (x * (14000 / x ^ 2)) = 16 (14000 / x)
4 sides = 64 (14000 / x) = 896000 / x
The total cost is:
y = floor + roof + walls
y = 4 (x ^ 2) + 3 (x ^ 2) + 896000 / x
y = 7 (x ^ 2) + 896000 / x
We derive the function:
y '= 14x - 896000 / x ^ 2
We match zero:
0 = 14x - 896000 / x ^ 2
We clear x:
14x = 896000 / x ^ 2
x ^ 3 = 896000/14
x = (896000/14) ^ (1/3)
x = 40
min cost (y) occurs when x = 40 ft
Then,
y = 7 * (40 ^ 2) + 896000/40
y = 33600 $
Then the height
z = 14000/40 ^ 2 = 8.75 ft
The price is:
floor = 4 * (40 ^ 2) = 6400
roof = 3 * (40 ^ 2) = 4800
walls = 16 * 4 * (40 * 8.75) = 22400
Total cost = $ 33600 (as calculated previously)
Answer:
The dimensions for minimum cost are:
40 * 40 * 8.75
