According to the National Telecommunication and Information Administration, 51% of U. S. households had Internet access in 2001. a. If we select four U. S. households at random (with replacement), what is the probability that all four had Internet access in 2001?
b. What is the probability that at least one of four randomly selected U. S. households had Internet access in 2001?

a

b

Step-by-step explanation:

From the question we are told that

   The probability of a US household having an internet access in 2001 is  p = 0.51

   The sample size is  n = 4

Generally the distribution of the number of US household with internet access at 2001  follows a binomial distribution  

i.e  

and the probability distribution function for binomial  distribution is  

Here C stands for combination hence we are going to be making use of the combination function in our calculators  

   Generally  the probability that all four had Internet access in 2001 is mathematically represented as

=>  

Generally  the probability that at least one of four randomly selected U.S. households had Internet access in 2001 is mathematically represented as

=>  

=>  

=>  

b. 6

step-by-step explanation:

we are given the table representing an exponential function.

we can see from the table that, as x is increasing, the value of y is increasing in the power of 6.

i.e. the following relation is obtained,

x = 1, y = 6

x = 2, y = 36 =

x = 3, y = 216 =

x = 4, y = 1296 =

x = 5, y = 7776 =

so, the geometric series obtained is .

thus, the common ratio is given by = = 6.

hence, the common ratio is 6.

ratio is 2 : 9

the answer is three

13.5 : 3