The area of the entire sector of DEF = 60 / 360 * PI * radius^2
sector area = 1 / 6 * 3.14159265... * 20^2
sector area =
<span>
<span>
<span>
209.4395102393
</span>
</span>
</span>
segment area = sector area - triangle DEF Area
triangle DEF Area = (1/2) * 20 * sine 60 * 20
triangle DEF Area = (1/2) * 0.86603 * 400
triangle DEF Area = <span><span><span>(1/2) * 346.412
</span>
</span>
</span>
triangle DEF Area =
<span>
<span>
<span>
173.206
</span>
</span>
</span>
segment area = <span>
<span>
209.4395102393
</span>
-173.206
</span>
segment area =
<span>
<span>
<span>
36.2335102393
</span>
</span>
</span>
segment area =
36.23 m
Source:
http://www.1728.org/circsect.htm
Step-by-step explanation:
the answer is above but there are data that not found
Answer:
the cost will be Rs 350
Step-by-step explanation:
The Scateboard cost Rs 450 each in the local store,The shop keeper says if I buy 1 I can buy another for only
of the normal price. We are asked to determine the cost of a second scateboard .
Hence the cost will be
of the normal cost that is


the cost will be Rs 350
Answer:
Step-by-step explanation:
Given that:
To bet $5 that the outcome is any one of these five possibilities: 0, 00, 1, 2, 3.
Let Y represent the Amount of net profit
Then, Y= {-5, 30}
The probability distribution of Y is:
Y -5 30
P(Y=y)

a) The expected value of X is given by:
![E[Y] =\sum y P(Y=y)= 30*\dfrac{5}{38}-5*\dfrac{33}{38}](https://tex.z-dn.net/?f=E%5BY%5D%20%3D%5Csum%20y%20P%28Y%3Dy%29%3D%2030%2A%5Cdfrac%7B5%7D%7B38%7D-5%2A%5Cdfrac%7B33%7D%7B38%7D)


b)
On a bet of $5 on the number 25 we are expected to loose 24 cents.
While on a $5 bet that the outcome is any one of the numbers 0,00, or 1 we are expected to loose 39 cents.
Hence, $5 bet on the number 27 is better. Because the expected loss is less in this bet
Answer:
C(0,6)
Step-by-step explanation:
1) find the length of the base of the triangle
AB = 7-3 = 4
2) find the the height of the triangle
A = (base x height)/2
A x 2 = base x height
height = (A x 2)/base
height = (12 x 2)/4 = 6
3) find the equation of the line that passes threw the points AB
the two points have the same y so the equation of the line is
y = 0
4) use the formula of the distance between a point and a line to find C
|ax+by+c|/√a^2 + b^2 = distance
x = 0
y = 6
C (0,6)