First find the area of the sector, which includes the shaded region's area and the area of the triangle JKL. Let
be the area of the sector. This area occurs in a fixed ratio with the area of the entire circle based on the measure of the central angle subtended by the arc LK:
![\dfrac A{54^\circ}=\dfrac{27^2\cdot3.14}{360^\circ}\implies A\approx343.359](https://tex.z-dn.net/?f=%5Cdfrac%20A%7B54%5E%5Ccirc%7D%3D%5Cdfrac%7B27%5E2%5Ccdot3.14%7D%7B360%5E%5Ccirc%7D%5Cimplies%20A%5Capprox343.359)
To get the area of the shaded region, subtract from
the area of the triangle.
The area of a triangle is 1/2 the base times the height. If we bisect the central angle with a line segment that meets the side KL at its midpoint, then we get a right triangle with hypotenuse 27 and one angle of measure 27º (half of the central angle). This triangle has base
and height
such that
![\sin27^\circ=\dfrac b{27}\implies b\approx12.258](https://tex.z-dn.net/?f=%5Csin27%5E%5Ccirc%3D%5Cdfrac%20b%7B27%7D%5Cimplies%20b%5Capprox12.258)
![\cos27^\circ=\dfrac h{27}\implies h\approx24.057](https://tex.z-dn.net/?f=%5Ccos27%5E%5Ccirc%3D%5Cdfrac%20h%7B27%7D%5Cimplies%20h%5Capprox24.057)
So the right triangle has area approximately 1/2*12.258*24.057, or about 147.443. Triangle JKL is made up of two of these right triangles, so it has area of 294.887.
Subtracting this from
gives an area of the shaded region of about 48.472, which we round up to 48.5.
X=2 is the correct answer x=2
Answer:
a) 0.82
b) 0.18
Step-by-step explanation:
We are given that
P(F)=0.69
P(R)=0.42
P(F and R)=0.29.
a)
P(course has a final exam or a research paper)=P(F or R)=?
P(F or R)=P(F)+P(R)- P(F and R)
P(F or R)=0.69+0.42-0.29
P(F or R)=1.11-0.29
P(F or R)=0.82.
Thus, the the probability that a course has a final exam or a research paper is 0.82.
b)
P( NEITHER of two requirements)=P(F' and R')=?
According to De Morgan's law
P(A' and B')=[P(A or B)]'
P(A' and B')=1-P(A or B)
P(A' and B')=1-0.82
P(A' and B')=0.18
Thus, the probability that a course has NEITHER of these two requirements is 0.18.