Divide 1 3/4 into 15. I like to change my fractions into decimals, so I will be using the decimal form of 1 3/4, which is 1.75
15 ÷ 1.75 = 8.571
Since there is still some left over, you would need to add one more roll. Therefore, you would need 9 rolls.
Answer:A
Step-by-step explanation:
Short answer
y >-x + 4
y <= 2(x - 1)^2 - 18
Comment
Let's establish what we are talking about. The question says purple and to me purple means the shaded area above the blue and to the upper right of the parabola.
Not so. It is the wine colored area right above the parabola according to the direction of he inequalities.
Second thing: the question is trying to see if you can read all he information you need from a graph.
Step One
Find the equation of the line.
We can read 2 points right away the x and y intercepts.
(4,0) is the x intercept
(0,4) is the y intercept
The general equation for the slope intercept equation is
y = mx + b So when x = 0 then b = y
4 = m*0 + b
b = 4
This far we have
y = mx + 4
when x = 4, y = 0
0 = m*4 + 4 Subtract 4 from both sides.
-4 = m*4 Divide by 4
-4/4 = m
m = - 1
Complete equation is y = -1x + 4
The inequality is y > -x + 4 Top answer block.
Step Two
Find the equation for the parabola
You could just read the minimum point which is (1,-18) There is a catch. (Isn't there always?) This may have a constant in front of the (x - a)^2 term. We'll check that out in the next step.
y = (x - 1)^2 - 18 is the basic vertex solution
Step Three
Find the value of a in
y = a(x - 1)^2 - 18
You have to do this because the answer has a space before the brackets. We need another point. The y intercept might work.
y = a(0 -1)^2 - 18
Use the point (0,-16)
-16 = a(1) - 18
-16 = a - 18 Add 18 to both sides.
2 = a
So the equation of the quadratic is
y <= 2(x - 1)^2 - 18 <<<<<< answer second section down on your question sheet.
Answer:
χ²R = 8.643
χ²L = 42.796
0.20 < σ < 0.45
Step-by-step explanation:
Given :
Sample size, n = 23
The degree of freedom, df = n - 1 = 23 - 1 = 22
At α - level = 99%
For χ²R ; 1 - (1 - 0.99)/2= 0.995 ; df = 22 ; χ²R = 8.643
For χ²L ; (1 - 0.99)/2 = 0.005 ; df = 22 ; χ²L = 42.796
The confidence interval of σ ;
s * √[(n-1)/χ²L] < σ < s * √[(n-1)/χ²R)]
0.28 * √(22/42.796) < σ < 0.28 * √(22/8.643)
0.2008 < σ < 0.4467
0.20 < σ < 0.45
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