Answer:
c
Step-by-step explanation:
I believe that number would be 6.
1/3n-12= -10
+12 +12
_________
1/3n = +2
x3 x3
_________
n = 6
Answer:
<u>Given function</u>
#15 Find the inverse of h(x)
<u>Substitute x with y and h(x) with x and solve for y:</u>
- x = 2y - 1
- 2y = x + 1
- y = 1/2x + 1/2
<u>The inverse is:</u>
#16 The graph with both lines is attached.
The x- and y-intercepts of both functions have reversed values.
#17 Table of the inverse function will contain same numbers with swapped domain and range.
<u>Initial look is like this:</u>
- <u>x | -3 | -2 | -1 | 0 | 1 | 2 | 3</u>
- h⁻¹(x) | -1 | | 0 | | 1 | | 2
<u>The rest of the table is filled in by finding the values:</u>
- <u>x | -3 | -2 | -1 | 0 | 1 | 2 | 3</u>
- h⁻¹(x) | -1 | -0.5 | 0 | 0.5 | 1 | 1.5 | 2
Answer:
a
Step-by-step explanation:
Recall that the "point-slope" form of a linear equation may be expressed as
y = mx + b,
where m is the gradient.
If m is negative, the gradient is negative.
If m is positive, the gradient is positive.
In our case, if we consider option A,
x + 3y = -2 (rearranging)
3y = -x -2
y = (-1/3) x - (2/3)
if we compare this to the general equation at the top, we can see that
gradient, = m = (-1/3) which is negative.
hence option a has a negative gradient.
Answer:
Option D
Step-by-step explanation:
A type I error occurs when you reject the null hypothesis when it is actually true.
The null hypothesis in this case is minimum breaking strength is less than or equal to 0.5.
A type one error would be allowing the production process to continue when the true breaking strength is below specifications.