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.
ANSWER: 31.25%
Explaination:
Suppose (x+y)
is the binomial distribution with 5 trials. Given that, probability of exactly two success in five trials of a binomial experiment in which the probability of success is 50%.
So, x = 0.5 and y = 0.5
The term
5C2*x²*y³ representing 2 successes and 3 failures.
Now find the value of this term.
5C2*(0.5)²*(0.5)³ = 
*
=10*0.03125
= 31.25%
Answer:
x=2 and y=0 is the required result.
Step-by-step explanation:
We have been given system of equations:
(1)
And 
(2)
We will use elimination method:
Multiply 1st equation by 3 and 2nd equation by 5 we get:
(3)
(4)
Now subtract (4) from (3) we get:
Now, put y=0 in (1) equation:
Hence, x=2 and y=0
Answer: the incubation time that separates the bottom 2.5% from the rest of the incubation times is 24.96 days.
Step-by-step explanation:
Suppose the incubation times are approximately normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = incubation times.
µ = mean time
σ = standard deviation
From the information given,
µ = 23 days
σ = 1 day
The probability value for the incubation time that separates the bottom 2.5% from the rest of the incubation times would be (1 - 2.5/100) = (1 - 0.025) = 0.975
Looking at the normal distribution table, the z score corresponding to the probability value is 1.96
Therefore,
1.96 = (x - 23)/1
x = 1.96 + 23
x = 24.96
