Answer:
Both are inverse pairs
Step-by-step explanation:
Question 11

(a) Rename g(x) as y

(b) Solve for x :

(c) Multiply each side by ⅝

(d) Switch x and y

(e) Rename y as the inverse function

(f) Compare with your function

f(x) and g(x) are inverse functions.
The graphs of inverse functions are reflections of each other across the line y = x.
In the first diagram, the graph of ƒ(x) (blue) is the reflection of g(x) (red) about the line y = x (black)
Question 12
h(x)= x - 2
(a) Rename h(x) as y
y = x - 2
(b) Solve for x:
x = y + 2
(c) Switch x and y
y = x + 2
(e) Rename y as the inverse function
h⁻¹(x) = x + 2
(f) Compare with your function
f(x) = x + 2
f(x) = h⁻¹(x)
h(x) and ƒ(x) are inverse functions.
The graph of h(x) (blue) reflects ƒ(x) (red) across the line y = x (black).
Answer:
C
Step-by-step explanation:
(x-3)+7
Answer:7/22
Sorry I don’t know how to explain.But here the answer.
Hope this helped!
Have a great day!
Answer:
Confidence Interval: (21596,46428)
Step-by-step explanation:
We are given the following data set:
10520, 56910, 52454, 17902, 25914, 56607, 21861, 25039, 25983, 46929
Formula:
where
are data points,
is the mean and n is the number of observations.


Sum of squares of differences = 551869365.6 + 524322983.6 + 340111052.4 + 259528878 + 65575984.41 + 510538544 + 147644370.8 + 80512934.41 + 64463235.21 + 166851472.4 = 2711418821

Confidence interval:

Putting the values, we get,

