Answer:
C
Step-by-step explanation:
Find the surface area:
Bottom base: 57 × 27 = 1,539
Top base: 21 × 27 = 567
Rectangular face: 23 × 27 = 621
Rectangular face: 23 × 27 = 621
Trapezoidal face: (21 + 57) ÷ 2 × 15 = 585
Trapezoidal face: (21 + 57) ÷ 2 × 15 = 585
1,539 + 567 + 621 + 621 + 585 + 585 = 4,518
The surface area is 4,518 square inches
Try plugging in 4,6, and -7 into the x's
Answer: x = 9/11
Step-by-step explanation:
First, distribute 9 to x and -1
9x - 9 = 2
Then add 9 to both sides
9x = 11
Then divide both sides by 9
x = 9/11
Hope it helps :)
Answer:
No, because it fails the vertical line test ⇒ B
Step-by-step explanation:
To check if the graph represents a function or not, use the vertical line test
<em>Vertical line test:</em> <em>Draw a vertical line to cuts the graph in different positions, </em>
- <em>if the line cuts the graph at just </em><em>one point in all positions</em><em>, then the graph </em><em>represents a function</em>
- <em>if the line cuts the graph at </em><em>more than one point</em><em> </em><em>in any position</em><em>, then the graph </em><em>does not represent a function </em>
In the given figure
→ Draw vertical line passes through points 2, 6, 7 to cuts the graph
∵ The vertical line at x = 2 cuts the graph at two points
∵ The vertical line at x = 6 cuts the graph at two points
∵ The vertical line at x = 7 cuts the graph at one point
→ That means the vertical line cuts the graph at more than 1 point
in some positions
∴ The graph does not represent a function because it fails the vertical
line test
Answer: 
Step-by-step explanation:
The confidence interval estimate for the population mean is given by :-
, where
is the sample mean and ME is the margin of error.
Given : Sample mean: 
The margin of error for a 98% confidence interval estimate for the population mean using the Student's t-distribution : 
Now, the confidence interval estimate for the population mean will be :-

Hence, the 98% confidence interval estimate for the population mean using the Student's t-distribution = 