Answer:
If y varies inversely as x, then we write it as:
y = k/x
where k is some constant.
If we multiply both sides of this equation by x we get an equation for k:
k = x*y
We are given that y=4 when x=7, therefore:
k = 7*4 = 28
Put this back into our original equation to get:
y = 28/x
Answer:
The probability is 
Step-by-step explanation:
If she has n distinct password candidates and only one of which will successfully log her into a secure system, the probability that her first first successful login will be on her k-th try is:
If k=1

Because, in her first try she has n possibles options and just one give her a successful login.
If k=2

Because, in her first try she has n possibles options and n-1 that are not correct, then, she has n-1 possibles options and 1 of that give her a successful login.
If k=3

Because, in her first try she has n possibles options and n-1 that are not correct, then, she has n-1 possibles options and n-2 that are not correct and after that, she has n-2 possibles options and 1 give her a successful login.
Finally, no matter what is the value of k, the probability that her first successful login will be (exactly) on her k-th try is 1/n
Answer:
any number that isn't 8 or above it
Step-by-step explanation:
Answer:
Part A = $180
Part B = $104
Step-by-step explanation:
Part A:
You do 255÷5=51
51 x 4 = 180
Part B:
You do 130÷5=26
26 x 4 = 104
Answer:
You must survey 784 air passengers.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence level of
, we have the following confidence interval of proportions.

In which
z is the zscore that has a pvalue of
.
The margin of error is:

95% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
Assume that nothing is known about the percentage of passengers who prefer aisle seats.
This means that
, which is when the largest sample size will be needed.
Within 3.5 percentage points of the true population percentage.
We have to find n for which M = 0.035. So






You must survey 784 air passengers.