Looking at this problem in terms of geometry makes it easier than trying to think of it algebraically.
If you want the largest possible x+y, it's equivalent to finding a rectangle with width x and length y that has the largest perimeter.
If you want the smallest possible x+y, it's equivalent to finding the rectangle with the smallest perimeter.
However, the area x*y must be constant and = 100.
We know that a square has the smallest perimeter to area ratio. This means that the smallest perimeter rectangle with area 100 is a square with side length 10. For this square, x+y = 20.
We also know that the further the rectangle stretches, the larger its perimeter to area ratio becomes. This means that a rectangle with side lengths 100 and 1 with an area of 100 has the largest perimeter. For this rectangle, x+y = 101.
So, the difference between the max and min values of x+y = 101 - 20 = 81.
Supplementary angles add to 180 so x + 128.7 = 180 subtract 128.7 from both sides; therefore, the answer is 51.3
I hope this helps!
D All real numbers Equal or greater that 2
Answer:
Step-by-step explanation:
For the null hypothesis,
µ = 60
For the alternative hypothesis,
h1: µ < 60
This is a left tailed test
Since the population standard deviation is not given, the distribution is a student's t.
Since n = 100,
Degrees of freedom, df = n - 1 = 100 - 1 = 99
t = (x - µ)/(s/√n)
Where
x = sample mean = 52
µ = population mean = 60
s = samples standard deviation = 22
t = (52 - 60)/(22/√100) = - 3.64
We would determine the p value using the t test calculator. It becomes
p = 0.00023
We would reject the null hypothesis if α = 0.05 > 0.00023
