Answer:
6 units
Step-by-step explanation:
Given: Points H and F lie on circle with center C. EG = 12, EC = 9 and ∠GEC = 90°.
To find: Length of GH.
Sol: EC = CH = 9 (Radius of the same circle are equal)
Now, GC = GH + CH
GC = GH + 9
Now In ΔEGC, using pythagoras theorem,
......(ΔEGC is a right triangle)
Now, Let GH = <em>x</em>
On rearranging,
So x = 6 and x = - 24
∵ x cannot be - 24 as it will not satisfy the property of right triangle.
Therefore, the length of line segment GH = 6 units. so, Option (D) is the correct answer.
Answer:
Cov(X, Y) =0.029.
Step-by-step explanation:
Given that :
The noise in a particular voltage signal has a constant mean of 0.9 V. that is μ = 0.9V ............(1)
Also, the two noise instances sampled τ seconds apart have a bivariate normal distribution with covariance.
0.04e–jτj/10 ............(2)
Having X and Y denoting the noise at times 3 s and 8 s, respectively, the difference of time = 8-3 = 5seconds.
That is, they are 5 seconds apart,
τ = 5 seconds..............(3)
Thus,
Cov(X, Y), for τ = 5seconds = 0.04e-5/10
= 0.04e-0.5 = 0.04/√e
= 0.04/1.6487
= 0.0292
Thus, Cov(X, Y) =0.029.
9.9^2X1.79
9.9^2=98.01
98.01X1.79=175.4379
Answer:
95% of the text messages have length between 23 units and 47 units.
Step-by-step explanation:
We are given the following in the question:
The lengths of text messages are normally distributed.
95% confidence interval:
(23,47)
Thus, we could interpret the confidence interval as:
About 95% of the text messages have length between 23 units and 47 units.
By Empirical rule for a normally distributed data, about 95% of data lies within 2 standard deviations of mean , thus we can write:
Thus, the mean length of text messages is 23 units and standard deviation is 6 units.
Hello,
Here is the demonstration in the book Person Guide to Mathematic by Khattar Dinesh.
Let's assume
P=cos(a)*cos(2a)*cos(3a)*....*cos(998a)*cos(999a)
Q=sin(a)*sin(2a)*sin(3a)*....*sin(998a)*sin(999a)
As sin x *cos x=sin (2x) /2
P*Q=1/2*sin(2a)*1/2sin(4a)*1/2*sin(6a)*....
*1/2* sin(2*998a)*1/2*sin(2*999a) (there are 999 factors)
= 1/(2^999) * sin(2a)*sin(4a)*...
*sin(998a)*sin(1000a)*sin(1002a)*....*sin(1996a)*sin(1998a)
as sin(x)=-sin(2pi-x) and 2pi=1999a
sin(1000a)=-sin(2pi-1000a)=-sin(1999a-1000a)=-sin(999a)
sin(1002a)=-sin(2pi-1002a)=-sin(1999a-1002a)=-sin(997a)
...
sin(1996a)=-sin(2pi-1996a)=-sin(1999a-1996a)=-sin(3a)
sin(1998a)=-sin(2pi-1998a)=-sin(1999a-1998a)=-sin(a)
So sin(2a)*sin(4a)*...
*sin(998a)*sin(1000a)*sin(1002a)*....*sin(1996a)*sin(1998a)
= sin(a)*sin(2a)*sin(3a)*....*sin(998)*sin(999) since there are 500 sign "-".
Thus
P*Q=1/2^999*Q or Q!=0 then
P=1/(2^999)
