Answer:
We know that the equation of the circle in standard form is equal to <em>(x-h)² + (y-k)² = r²</em> where (h,k) is the center of the circle and r is the radius of the circle.
We have x² + y² + 8x + 22y + 37 = 0, let's get to the standard form :
1 - We first group terms with the same variable :
(x²+8x) + (y²+22y) + 37 = 0
2 - We then move the constant to the opposite side of the equation (don't forget to change the sign !)
(x²+8x) + (y²+22y) = - 37
3 - Do you recall the quadratic identities ? (a+b)² = a² + 2ab + b². Now that's what we are trying to find. We call this process <u><em>"completing the square"</em></u>.
x²+8x = (x²+8x + 4²) - 4² = (x+4)² - 4²
y²+22y = (y²+22y+11²)-11² = (y+11)²-11²
4 - We plug the new values inside our equation :
(x+4)² - 4² + (y+22)² - 11² = -37
(x+4)² + (y+22)² = -37+4²+11²
(x+4)²+(y+22)² = 100
5 - We re-write in standard form :
(x-(-4)²)² + (y - (-22))² = 10²
And now it is easy to identify h and k, h = -4 and k = - 22 and the radius r equal 10. You can now complete the sentence :)
Answer:
no of sample within two percentage points is 5409
no of sample not use prior estimates is 8321
Step-by-step explanation:
Given data
E = 2% = 0.02
male m = 21.2 % = 0.212
female f = 19.7 % = 0.197
to find out
no of sample within two percentage points and if does not use any prior estimates
solution
first we calculate no of sample within two percentage points
we know Z critical value for 99% is 2.58
so no of sample will be
n = (Z / E )² × m(1-m) + f ( 1-f )
put all the value we get no of sample
n = (2.58 / 0.02 )² × 0.212 ( 1 - 0.212 ) + 0.197 ( 1 - 0.197 )
n = 16641 ×0.3250
no of sample within two percentage points is 5409
in 2nd part no of sample not use prior estimates
here than f and m will be 0.5
than
n = (Z / E )² × m(1-m) + f ( 1-f )
put all value
n = (2.58 / 0.02 )² × 0.5 ( 1 - 0.5 ) + 0.5 ( 1 - 0.5 )
n = 16641 × 0.5
no of sample not use prior estimates is 8321
Answer:
n=-4
Step-by-step explanation:
n/2+5=3
n/2=-2
n=-4
Y=-13/2x+43/2 I hope this helped :3
It is 16.256 because I went to google and found it.... yep that simple.
