Answer:
The concentration of salt in the tank approaches 
Step-by-step explanation:
Data provide in the question:
Water contained in the tank = 8000 L
Salt per litre contained in Brine = 35 g/L
Rate of pumping water into the tank = 25 L/min
Concentration of salt 
Now,
Dividing both numerator and denominator by 
, we have
Here,
The concentration of salt in the tank approaches
Minus 3y both sides
2y-6=-20
add 6 to both sids
2y=-14
divide 2
y=-7
(a) 0.059582148 probability of exactly 3 defective out of 20
(b) 0.98598125 probability that at least 5 need to be tested to find 2 defective.
(a) For exactly 3 defective computers, we need to find the calculate the probability of 3 defective computers with 17 good computers, and then multiply by the number of ways we could arrange those computers. So
0.05^3 * (1 - 0.05)^(20-3) * 20! / (3!(20-3)!)
= 0.05^3 * 0.95^17 * 20! / (3!17!)
= 0.05^3 * 0.95^17 * 20*19*18*17! / (3!17!)
= 0.05^3 * 0.95^17 * 20*19*18 / (1*2*3)
= 0.05^3 * 0.95^17 * 20*19*(2*3*3) / (2*3)
= 0.05^3 * 0.95^17 * 20*19*3
= 0.000125* 0.418120335 * 1140
= 0.059582148
(b) For this problem, let's recast the problem into "What's the probability of having only 0 or 1 defective computers out of 4?" After all, if at most 1 defective computers have been found, then a fifth computer would need to be tested in order to attempt to find another defective computer. So the probability of getting 0 defective computers out of 4 is (1-0.05)^4 = 0.95^4 = 0.81450625.
The probability of getting exactly 1 defective computer out of 4 is 0.05*(1-0.05)^3*4!/(1!(4-1)!)
= 0.05*0.95^3*24/(1!3!)
= 0.05*0.857375*24/6
= 0.171475
So the probability of getting only 0 or 1 defective computers out of the 1st 4 is 0.81450625 + 0.171475 = 0.98598125 which is also the probability that at least 5 computers need to be tested.
Answer:
81%
Step-by-step explanation:
Let 'L' be the dominant and 'l' e the recessive allele for ‘lazybuttness’.
Since ‘lazybuttness’ is an autosomal dominant condition, the 19% of students affected by the condition correspond to the homozygous dominant (LL) and heterozygous (Ll) genotypes. Therefore, the rest of the population has the homozygous recessive genotype (ll) and is not affected. The frequency of students not affected is:
F = 100% - 19% = 81%
To add the two expressions, we can write it as:
4.6x-3 + (-5.3x+9)
We can distribute the plus sign (which means just drop the parenthesis in this case):
4.6x-3-5.3x+9
Now, we can simplify by combining like terms:
-0.7x+6
