Answer:
0.621
Step-by-step explanation:
Number of people that have run a red light in the last year = 290
Number of people that have not run a red light in the last year = 177
Total number of respondents = 290 + 177
= 467
Pr(if a person is chosen at random, they have run a red light in the last year) = 
= 
= 0.621
The probability that if a person is chosen at random, they have run a red light in the last year is 0.621.
Answer:
the probability that a code word contains exactly one zero is 0.0064 (0.64%)
Step-by-step explanation:
Since each bit is independent from the others , then the random variable X= number of 0 s in the code word follows a binomial distribution, where
p(X)= n!/((n-x)!*x!*p^x*(1-p)^(n-x)
where
n= number of independent bits=5
x= number of 0 s
p= probability that a bit is 0 = 0.8
then for x=1
p(1) = n*p*(1-p)^(n-1) = 5*0.8*0.2^4 = 0.0064 (0.64%)
therefore the probability that a code word contains exactly one zero is 0.0064 (0.64%)
Answer:
2. y = x + 16; 0.21y = 0.27x + 2.88
Step-by-step explanation:
You have two conditions:
- Volume of 18 % + volume of 27 % = volume of 21 %
- Mass of acid in 18 % + mass of acid in 27% = mass of acid in 21 %
For the first condition,
16 + x = y Transpose
(1) y = x + 16
=====
For the second condition
V₁c₁ + V₂c₂ = V₃c₃
16×0.18 + 0.27x = 0.21y
2.88 + 0.27x = 0.21y Transpose
(2) 0.21y = 0.27x + 2.88
The system of equations is y = x + 16; 0.21y = 0.27x + 2.88
Answer:
He payed off 6,800$
Step-by-step explanation:
