Q: 115 z 1

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About 15.87% of people have an IQ over 115.z=(115-100)/15=1Z-score of 1 on a normal distribution means 84.13% have an IQ below 115 and 15.87% have an IQ above 115.IQ Graphhttp://8.17.172.75/uiq/images/bellcurve.gif

sin(z)= (e^(i*z)-e^(-i*z))/(2*i) where i=(-1)^(1/2)

Z2 - 5Z + 4 = 0 (Z- 1)(Z - 4) so, Z = 1 ---------- and Z = 4 -----------

y * (x-1) = z Express as x,divide both sides by y(x - 1) = z/yadd 1 on both sidesx = z/y + 1

(1/x) - (1/y) = (1/z) Get the left-hand side over a common denominator:- (y-x)/xy = 1/z Take the reciprocal of both sides:- z = xy / (y-x)

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About 15.87% of people have an IQ over 115.z=(115-100)/15=1Z-score of 1 on a normal distribution means 84.13% have an IQ below 115 and 15.87% have an IQ above 115.IQ Graphhttp://8.17.172.75/uiq/images/bellcurve.gif

Z + or - i and z-1

Let the sides be x y z and their opposite angles be X Y Z Using the cosine rule angle X = 41 degrees Using the cosine rule angle Y = 115 degrees Angle Z: 180-41-115 = 24 degrees

If ' z ' is greater than ' 1 ', then ' z ' is. If ' z ' is less than ' 1 ', then ' 1 ' is.

z2 = z * z * 1; z = z * 1. Greatest common factor is z.

jay z is worth over 400 million and eminem is abour 115 million

Pierre De Fermat 's last Theorem. The conditions: x,y,z,n are the integers and >0. n>2. Proof: z^n=/x^n+y^n. We have; z^3=[z(z+1)/2]^2-[(z-1)z/2]^2 Example; 5^3=[5(5+1)/2]^2-[5(5-1)/2]^2=225-100=125 And z^3+(z-1)^3=[z(z+1)/2]^2-[(z-2)(z-1)/2]^2 Example; 5^3+4^3=[5(5+1)/2]^2-[(5-2)(5-1)/2]^2=225-36=189 And z^3+(z-1)^3+(z-2)^3=[z(z+1)/2]^2-[(z-3)(z-2)/2]^2 Example 5^3+4^3+2^3=[5(5+1)/2]^2-[(5-3)(5-2)/2]^2=225-9=216 And z^3+(z-1)^3+(z-2)^3+(z-3)^3=[z(z+1)/2]^2-[(z-4)(z-3)/2]^2 Example 5^3+4^3+3^3+2^3=[5(5+1)/2]^2-[(5-4)(5-3)/2]^2=225-1=224 General: z^3+(z-1)^3+....+(z-m)^3=[z(z+1)/2]^2-[(z-m-1)(z-m)/2]^2 We have; z^3=z^3+(z-m-1)^3 - (z-m-1)^3. Because: z^3+(z-m-1)^3=[z^3+(z-1)^3+....+(z-m-1)^3] - [(z-1)^3+....+(z-m)^3] So z^3=[z(z+1)/2]^2-[(z-m-2)(z-m-1)/2]^2 - [z(z-1)/2]^3+[(z-m-1)(z-m)/2]^2 - (z-m-1)^3. Similar: z^3=z^3+(z-m-2)^3 - (z-m-2)^3. So z^3=[z(z+1)/2]^2-[(z-m-3)(z-m-2)/2]^2 - [z(z-1)/2]^3+[(z-m-2)(z-m-1)/2]^2 - (z-m-2)^3. .... .... Suppose: z^n=x^n+y^n So z^(n-3)*z^3=x^(n-3)^n*x^3+y^(n-3)*y^3. So z^(n-3)*{[z(z+1)/2]^2-[(z-m-2)(z-m-1)/2]^2 - [z(z-1)/2]^3+[(z-m-1)(z-m)/2]^2 - (z-m-1)^3}=x^(n-3)*{[x(x+1)/2]^2-[(x-m-2)(x-m-1)/2]^2 - [x(x-1)/2]^3+[(x-m-1)(x-m)/2]^2 - (x-m-1)^3}+y^(n-3)*{[y(y+1)/2]^2-[(y-m-2)(y-m-1)/2]^2 - [y(y-1)/2]^3+[(y-m-1)(y-m)/2]^2 - (y-m-1)^3} Similar: z^(n-3)*{[z(z+1)/2]^2-[(z-m-3)(z-m-2)/2]^2 - [z(z-1)/2]^3+[(z-m-2)(z-m-1)/2]^2 - (z-m-2)^3=x^(n-3)*{[x(x+1)/2]^2-[(x-m-3)(x-m-2)/2]^2 - [x(x-1)/2]^3+[(x-m-2)(x-m-1)/2]^2 - (x-m-2)^3+y^(n-3)*{[y(y+1)/2]^2-[(y-m-3)(y-m-2)/2]^2 - [y(y-1)/2]^3+[(y-m-2)(y-m-1)/2]^2 - (y-m-2)^3. .... .... Because it is codified . So Impossible all are the integers. So: z^n=/x^n+y^n. ISHTAR.

Pierre De Fermat 's last Theorem. The conditions: x,y,z,n are the integers and >0. n>2. Proof: z^n=/x^n+y^n. We have; z^3=[z(z+1)/2]^2-[(z-1)z/2]^2 Example; 5^3=[5(5+1)/2]^2-[5(5-1)/2]^2=225-100=125 And z^3+(z-1)^3=[z(z+1)/2]^2-[(z-2)(z-1)/2]^2 Example; 5^3+4^3=[5(5+1)/2]^2-[(5-2)(5-1)/2]^2=225-36=189 And z^3+(z-1)^3+(z-2)^3=[z(z+1)/2]^2-[(z-3)(z-2)/2]^2 Example 5^3+4^3+2^3=[5(5+1)/2]^2-[(5-3)(5-2)/2]^2=225-9=216 And z^3+(z-1)^3+(z-2)^3+(z-3)^3=[z(z+1)/2]^2-[(z-4)(z-3)/2]^2 Example 5^3+4^3+3^3+2^3=[5(5+1)/2]^2-[(5-4)(5-3)/2]^2=225-1=224 General: z^3+(z-1)^3+....+(z-m)^3=[z(z+1)/2]^2-[(z-m-1)(z-m)/2]^2 We have; z^3=z^3+(z-m-1)^3 - (z-m-1)^3. Because: z^3+(z-m-1)^3=[z^3+(z-1)^3+....+(z-m-1)^3] - [(z-1)^3+....+(z-m)^3] So z^3=[z(z+1)/2]^2-[(z-m-2)(z-m-1)/2]^2 - [z(z-1)/2]^3+[(z-m-1)(z-m)/2]^2 - (z-m-1)^3. Similar: z^3=z^3+(z-m-2)^3 - (z-m-2)^3. So z^3=[z(z+1)/2]^2-[(z-m-3)(z-m-2)/2]^2 - [z(z-1)/2]^3+[(z-m-2)(z-m-1)/2]^2 - (z-m-2)^3. .... .... Suppose: z^n=x^n+y^n So z^(n-3)*z^3=x^(n-3)^n*x^3+y^(n-3)*y^3. So z^(n-3)*{[z(z+1)/2]^2-[(z-m-2)(z-m-1)/2]^2 - [z(z-1)/2]^3+[(z-m-1)(z-m)/2]^2 - (z-m-1)^3}=x^(n-3)*{[x(x+1)/2]^2-[(x-m-2)(x-m-1)/2]^2 - [x(x-1)/2]^3+[(x-m-1)(x-m)/2]^2 - (x-m-1)^3}+y^(n-3)*{[y(y+1)/2]^2-[(y-m-2)(y-m-1)/2]^2 - [y(y-1)/2]^3+[(y-m-1)(y-m)/2]^2 - (y-m-1)^3} Similar: z^(n-3)*{[z(z+1)/2]^2-[(z-m-3)(z-m-2)/2]^2 - [z(z-1)/2]^3+[(z-m-2)(z-m-1)/2]^2 - (z-m-2)^3=x^(n-3)*{[x(x+1)/2]^2-[(x-m-3)(x-m-2)/2]^2 - [x(x-1)/2]^3+[(x-m-2)(x-m-1)/2]^2 - (x-m-2)^3+y^(n-3)*{[y(y+1)/2]^2-[(y-m-3)(y-m-2)/2]^2 - [y(y-1)/2]^3+[(y-m-2)(y-m-1)/2]^2 - (y-m-2)^3. .... .... Because it is codified . So Impossible all are the integers. So: z^n=/x^n+y^n. ISHTAR.

Pierre De Fermat 's last Theorem. The conditions: x,y,z,n are the integers and >0. n>2. Proof: z^n=/x^n+y^n. We have; z^3=[z(z+1)/2]^2-[(z-1)z/2]^2 Example; 5^3=[5(5+1)/2]^2-[5(5-1)/2]^2=225-100=125 And z^3+(z-1)^3=[z(z+1)/2]^2-[(z-2)(z-1)/2]^2 Example; 5^3+4^3=[5(5+1)/2]^2-[(5-2)(5-1)/2]^2=225-36=189 And z^3+(z-1)^3+(z-2)^3=[z(z+1)/2]^2-[(z-3)(z-2)/2]^2 Example 5^3+4^3+2^3=[5(5+1)/2]^2-[(5-3)(5-2)/2]^2=225-9=216 And z^3+(z-1)^3+(z-2)^3+(z-3)^3=[z(z+1)/2]^2-[(z-4)(z-3)/2]^2 Example 5^3+4^3+3^3+2^3=[5(5+1)/2]^2-[(5-4)(5-3)/2]^2=225-1=224 General: z^3+(z-1)^3+....+(z-m)^3=[z(z+1)/2]^2-[(z-m-1)(z-m)/2]^2 We have; z^3=z^3+(z-m-1)^3 - (z-m-1)^3. Because: z^3+(z-m-1)^3=[z^3+(z-1)^3+....+(z-m-1)^3] - [(z-1)^3+....+(z-m)^3] So z^3=[z(z+1)/2]^2-[(z-m-2)(z-m-1)/2]^2 - [z(z-1)/2]^3+[(z-m-1)(z-m)/2]^2 - (z-m-1)^3. Similar: z^3=z^3+(z-m-2)^3 - (z-m-2)^3. So z^3=[z(z+1)/2]^2-[(z-m-3)(z-m-2)/2]^2 - [z(z-1)/2]^3+[(z-m-2)(z-m-1)/2]^2 - (z-m-2)^3. .... .... Suppose: z^n=x^n+y^n So z^(n-3)*z^3=x^(n-3)^n*x^3+y^(n-3)*y^3. So z^(n-3)*{[z(z+1)/2]^2-[(z-m-2)(z-m-1)/2]^2 - [z(z-1)/2]^3+[(z-m-1)(z-m)/2]^2 - (z-m-1)^3}=x^(n-3)*{[x(x+1)/2]^2-[(x-m-2)(x-m-1)/2]^2 - [x(x-1)/2]^3+[(x-m-1)(x-m)/2]^2 - (x-m-1)^3}+y^(n-3)*{[y(y+1)/2]^2-[(y-m-2)(y-m-1)/2]^2 - [y(y-1)/2]^3+[(y-m-1)(y-m)/2]^2 - (y-m-1)^3} Similar: z^(n-3)*{[z(z+1)/2]^2-[(z-m-3)(z-m-2)/2]^2 - [z(z-1)/2]^3+[(z-m-2)(z-m-1)/2]^2 - (z-m-2)^3=x^(n-3)*{[x(x+1)/2]^2-[(x-m-3)(x-m-2)/2]^2 - [x(x-1)/2]^3+[(x-m-2)(x-m-1)/2]^2 - (x-m-2)^3+y^(n-3)*{[y(y+1)/2]^2-[(y-m-3)(y-m-2)/2]^2 - [y(y-1)/2]^3+[(y-m-2)(y-m-1)/2]^2 - (y-m-2)^3. .... .... Because it is codified . So Impossible all are the integers. So: z^n=/x^n+y^n. ISHTAR.

120 is one standard deviation greater than the mean (z = 1).So you want Pr(z < 1)The probability of a score at most 1 sd away from the mean is 0.68That is Pr(-1 < z < 1) = 0.68So Pr(|z| > 1) = 1 - 0.68 = 0.32then by symmetry,Pr(z > 1) = 1/2*0.32 = 0.16So Pr(z < 1) = 1 - 0.16 = 0.84120 is one standard deviation greater than the mean (z = 1).So you want Pr(z < 1)The probability of a score at most 1 sd away from the mean is 0.68That is Pr(-1 < z < 1) = 0.68So Pr(|z| > 1) = 1 - 0.68 = 0.32then by symmetry,Pr(z > 1) = 1/2*0.32 = 0.16So Pr(z < 1) = 1 - 0.16 = 0.84120 is one standard deviation greater than the mean (z = 1).So you want Pr(z < 1)The probability of a score at most 1 sd away from the mean is 0.68That is Pr(-1 < z < 1) = 0.68So Pr(|z| > 1) = 1 - 0.68 = 0.32then by symmetry,Pr(z > 1) = 1/2*0.32 = 0.16So Pr(z < 1) = 1 - 0.16 = 0.84120 is one standard deviation greater than the mean (z = 1).So you want Pr(z < 1)The probability of a score at most 1 sd away from the mean is 0.68That is Pr(-1 < z < 1) = 0.68So Pr(|z| > 1) = 1 - 0.68 = 0.32then by symmetry,Pr(z > 1) = 1/2*0.32 = 0.16So Pr(z < 1) = 1 - 0.16 = 0.84

Z is a variable with mean 0 and variance 1.Z is a variable with mean 0 and variance 1.Z is a variable with mean 0 and variance 1.Z is a variable with mean 0 and variance 1.

x + 1 = y y + 3 = z z = y + 3 = (x + 1) + 3 = x + 4 Or: x = y - 1 = (z - 3) - 1 = z - 4 Which results in the same: x exceeds z by 4.