Answer:
(the statement does not appear to be true)
Step-by-step explanation:
I don't think the statement is true, but you CAN compute the intercepted arc from the angle.
Note that BFDG is a convex quadrilateral, so its angles sum to 360. Since we know the inscribed circle touches the angle tangentially, angles BFD and BGD are both right angles, with a measure of 90 degrees.
Therefore, adding the angles together, we have:
alpha + 90 + 90 + <FDG = 360
Therefore, <FDG, the inscribed angle, is 180-alpha (ie, supplementary to alpha)
The answer: 40% of 60 is 24
Answer:
Option C: 41
Step-by-step explanation:
<ABD+<CBD=90
<ABD=8x+1
<CBD=6x+5
(8x+1)+(6x+5)=90
Combine like terms
(8x+6x)+(1+5)=90
14x+6=90
14x=90-6
14x=84
divide both sides by 14
x=6
Now we plug that value into <DBC = 6x+5
6(6)+5=
36+5
41
3 squares = 4 circles, so (number of squares)/(number of circles) = 3/4.
3/4 = 12/16
:::::
4 squares = 2 circles, so (number of squares)/(number of circles) = 4/2.
4/2 = 2/1
:::::
2 squares = 5 circles, so (number of squares)/(number of circles) = 2/5.
2/5 = 4/10
Let X be the number of burglaries in a week. X follows Poisson distribution with mean of 1.9
We have to find the probability that in a randomly selected week the number of burglaries is at least three.
P(X ≥ 3 ) = P(X =3) + P(X=4) + P(X=5) + ........
= 1 - P(X < 3)
= 1 - [ P(X=2) + P(X=1) + P(X=0)]
The Poisson probability at X=k is given by
P(X=k) =
Using this formula probability of X=2,1,0 with mean = 1.9 is
P(X=2) =
P(X=2) =
P(X=2) = 0.2698
P(X=1) =
P(X=1) =
P(X=1) = 0.2841
P(X=0) =
P(X=0) =
P(X=0) = 0.1495
The probability that at least three will become
P(X ≥ 3 ) = 1 - [ P(X=2) + P(X=1) + P(X=0)]
= 1 - [0.2698 + 0.2841 + 0.1495]
= 1 - 0.7034
P(X ≥ 3 ) = 0.2966
The probability that in a randomly selected week the number of burglaries is at least three is 0.2966