Solve compound inequalities<span> in the form of or and express the </span>solution<span> graphically. ... determining the</span>solution to the compound inequality<span>, as in the example </span>below<span>. .... </span>Which of the following compound inequalities<span> represents the graph
</span>
Since the skier has to rent skis for 30 dollars per day with
both passes, we can ignore it while solving the question.
A season pass is 350, while a daily pass is 75.

So a seasonal pass is equivalent to having 4.6667 daily passes. And you can't buy 0.6667 of a pass.
So, if you went at least 5 days with a seasonal pass, then the seasonal pass would be less expensive than a daily pass.
The answer is C
Insert y=X-2 into the equation y=-X+2
You will get the point (2,0)
Insert y=2X-1 into the same equation
you will get the point (1,1).
Without solving for the last point
you will realize that your answer is C
Note: This approach is appropriate for solving objective questions
The supplementary angle would be 180 degrees, and the other angle would be 92.2 degrees
Answer:
The 95% confidence interval is "87.94, 91.86".
Step-by-step explanation:
The given values are:
Sample batches,
n = 25
Sample mean concentration,
= 89.9 mg
Amount of chemical,
σ = 5
α = 0.05
The critical value from Z table will be:
= 
= 
= 
Now,
The confidence interval will be:
= 
On substituting the values, we get
= 
= 
= 
Lower limit will be:
= 
= 
Upper limit will be:
= 
= 