The total cost would  be:
$20 = two people- 2*20=40$
20 = guest
20*20=
400 would be the cost
 
        
             
        
        
        
Answer:
A sample size of 345 is needed so that the confidence interval will have a margin of error of 0.07
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
, and a confidence level of  , we have the following confidence interval of proportions.
, we have the following confidence interval of proportions.

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

In this problem, we have that:

99.5% confidence level
So  , z is the value of Z that has a pvalue of
, z is the value of Z that has a pvalue of  , so
, so  .
.
Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.07?
This is n when M = 0.07. So







A sample size of 345 is needed so that the confidence interval will have a margin of error of 0.07
 
        
             
        
        
        
The first integer is going to be x and the second one is going to be y.
We now turn the clues into equations:
x = y - 5
xy = 84
The substitution method:
(y - 5)y = 84
y^2 - 5y - 84 = 0
(y+7)(y-12) = 0
y = -7, 12.
If y = -7, then x = -12.
If y = 12, then x = 7
 
        
             
        
        
        
In 52-card deck there are 36 numbered card and 52/4=13 spade card
 so probability that you are dealt a  numbered card is 36/52 =9/13  a spade card is 13/52=1/4