Least to greatest, the triples are (2x, x²-1, x²+1). Thus the value in the first column of the table is half the value of the first of the triples. Your completed table will be

-3y + 5 = -4 Subtract 5 from both sides
-3y = -9 Divide both sides by -3
y = 3
Now, plug that into 5y
5y Plug in 3
5(3) Multiply
15
5y = 15
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)
Answer:
c
Step-by-step explanation:
bc it has the most 5's in it
If a random sample of 20 persons weighed 3,460, the sample mean x-bar would be 3460/20 = 173 pounds.
The z-score for 173 pounds is given by:

Referring to a standard normal distribution table, and using z = 0.66, we find:

Therefore

The answer is: 0.2546