Answer:
Stratified sampling technique(A)
Step-by-step explanation:
From the question, the population of an high school from which selection was made equals 461 sophomores, 328 juniors and 558 seniors.
35 sophomores, 69 juniors and 24 seniors are randomly selected. The technique used in selecting is Stratified sampling technique. This is because stratified sampling involves dividing the entire population into stratas and then selects a final sample randomly from the different strata. This means that a smaller part of the entire population is used as a sample in drawing conclusions for the entire population.
Answer:
The constant would be A) 9
Step-by-step explanation:
The Constant in a math expression is One number that is on its own or without a varialbe
Answer:
8x² - 15y² + xy
Step-by-step explanation:
(4x + 5y) (2x - 3y) + 3xy
multiplying the terms in brackets
(4x) (2x - 3y) + (5y) (2x - 3y) + 3 xy
multiplying with each terms inside the bracket
(4x)(2x) - (4x) (3y) + (5y) (2x) - (5y) (3y) + 3xy
doing the product each of the pair of terms
8x² - 12xy + 10xy - 15y² + 3xy
taking the sum of terms with coefficient "xy"
8x² - 15y² -2xy + 3xy
8x² - 15y² + xy
Answer:
1060934.37624
Step-by-step explanation:
Nice troll question
Answer:
the chemist should use 60 liters of 55% solution and 40 litres of 30% solution in order to prepare 100 liters of 45% purity of sulphuric acid.
Step-by-step explanation:
From the given information,
Let x be the litres of 55% pure solution
Let y be the litres of 30% pure solution
Also;
Given that our total volume of solution is 100 litres
x+y =100 ---- (1)
The total solution of pure by related by the sum of the individual pure concentrations to make up the concentration of final solution.
(0.55)(x)+(0.30)(y) = 0.45(100) ---- (2)
From equation (1)
Let ; y = 100 - x
Replacing the value for y = 100 - x into equation (2)
(0.55)(x)+(0.30)(100-x) = 0.45(100)
0.55x + 30 - 0.30x = 45
0.55x - 0.30x = 45 - 30
0.25x = 15
x = 15/0.25
x = 60 liters of 55% solution
From ; y = 100 - x
y = 100 - 60
y = 40 litres of 30% solution.
Therefore, the chemist should use 60 liters of 55% solution and 40 litres of 30% solution in order to prepare 100 liters of 45% purity of sulphuric acid.
