Given:
Planes X and Y are perpendicular to each other
Points A, E, F, and G are points only in plane X
Points R and S are points in both planes X and Y
Lines EA and FG are parallel
The lines which could be perpendicular to RS are EA and FG.
Answer: The mean number of checks written per day 
Standard deviation
Variance 
Step-by-step explanation:
Given : The total number of checks wrote by person in a year = 126
Assume that the year is not a leap year.
Then 1 year = 365 days
Let the random variable x represent the number of checks he wrote in one day.
Then , the mean number of checks wrote by person each days id=s given by :-

Since , the distribution is Poisson distribution , then the variance must equal to the mean value i.e. 
Standard deviation : 
When each x value has only one y value, the relation is a function

The shown pair of angles are Co interior angle pair, therefore the sum of those angles is 180°






The required value of x is 7°
<span>Determine the lower limit of the 95% interval for the true mean weight. ... A diver descended at a constant rate of 12.24 feet every 3 minutes. .... weight of 16.05 ounces. at the 5% level of significance can we conclude that ...</span><span>
</span>