Answer:
Every ordered pair within this region will satisfy the inequality y ≥ x. Incorrect. The boundary line here is y = x, and the correct region is shaded, but remember that a dotted line is used for < and >. The inequality you are graphing is y ≥ x, so the boundary line should be solid.
We can use point slope form to solve for this.
y - 1 = -2(x - 2)
Simplify.
y - 1 = -2x + 4
Add 1 to both sides.
y = -2x + 5
Now, we can input -3 for x and solve for y, or r in this case.
y = -2(-3) + 5
y = 6 + 5
y = 11
r = 11
Start at one of the vertices and draw a line from there to a point on the oposite side, not ending at another vertex
Step-by-step explanation:
Population Mean (u) = 3.50
Sample (n)= 36
Sample mean (x) = 3.60
Population standard deviation (s)= 0.40
Test statistics:
(Null hypothesis) H0: u= 3.5 (Population mean is equal to 3.5)
(Alternate hypothesis) H1: u> 3.5 (Population mean is greater than 3.5)
Z= 
= 
= 
= 1.5
critical value= Z0.05= 1.645 (From Z table)
Since, Z value is less than critical Z value that is Z<1.645
We cannot reject null hypothesis
So, we decide to reject that the mean GPA of graduates exceeds 3.50
Answer:
(f + g)(x) = I2x + 1I + 1 ⇒ C
Step-by-step explanation:
Let us solve the question
∵ f(x) = I2x + 1I + 3
∵ g(x) = -2
→ We need to find (f + g)(x), which means add the two functions
∵ (f + g)(x) = f(x) + g(x)
→ Substitute the right side of each function on the right side
∴ (f + g)(x) = I2x + 1I + 3 + (-2)
→ Remember (+)(-) = (-)
∴ (f + g)(x) = I2x + 1I + 3 - 2
→ Add the like terms in the right side
∵ (f + g)(x) = I2x + 1I + (3 - 2)
∴ (f + g)(x) = I2x + 1I + 1
