The area of a polygon = n*side^2 / (4 * tan(180/n))
where "n" is the number of sides
Let's calculate area for side = 2
area = 8*2^2 / (4 * tan(180/8))
area = 32 / (4 * tan(22.5))
area = 32 / 4*0.41421
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area = 19.3138746047
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Now, let's calculate area for side = 6
area = 8*6^2 / (4 * tan(180/8))
area = 288 /
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1.65684
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area =
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173.824871442
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173.824871442
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/ </span><span><span>19.3138746047 = </span> 9
So, the area would be 9 times larger.
ALSO, looking at the formula
</span>n*side^2 / (4 * tan(180/n))
we can see that the side length appears just once in the formula and we are to square it in the calculation. So, if we increase the side length is increased by 3, then the area increases by 3^2 or 9.
< means “Less than” and the other one means “less than or equal to”
Answer:
Please find attached the required graph of the equation, y = x² - 2·x - 8, created using MS Excel
Step-by-step explanation:
The given function required to be plot is y = x² - 2·x - 8
Therefore, the coordinates of the vertex of the parabola formed by the equation (h, k) is given as follows;
h = -(-2)/(2×1) = 1
k = 1² - 2×1 - 8 = -9
The coordinates of the vertex = (1, -9)
The roots of the equation is given when y = 0, as follows;
At the roots, we have;
x² - 2·x - 8 = 0
By factorizing, we get
x² - 2·x - 8 = (x - 4)·(x + 2) = 0
Therefore. the roots of the equation are;
x = 4, and x = -2
The coordinates of the roots are;
(4, 0), and (-2, 0)
Two other points can be found at when x = 3 and when x = 0 as follows;
When x = 3, we have;
y = 3² - 2×3 - 8 = -5
(3, -5)
When x = 0, we have;
y = 0² - 2×0 - 8 = -8
(0, -8)
The five points are;
(1, -9), (4, 0), (-2, 0), (3, -5), (0, -8)
The graph of the equation created using MS Excel showing the five points is attached.
Answer: (0.120,0.160)
Step-by-step explanation:
Given : Sample size : 
Number of disks were not defective =701
Then , the number of disks which are defective = 
Now, the proportion of disks which are defective : 
Significance level : 
Critical value : 
The confidence interval for population proportion is given by :-

Hence, the 90% confidence interval for the population proportion of disks which are defective = (0.120,0.160)