Best Answer

Assuming that these refer to the standard normal distribution, the areas are as follows:

z = 0 to 1.35: 0.411

z = 0 to 2.24: 0.487

z = 1.35 to 2.24: 0.076

Assuming that these refer to the standard normal distribution, the areas are as follows:

z = 0 to 1.35: 0.411

z = 0 to 2.24: 0.487

z = 1.35 to 2.24: 0.076

Assuming that these refer to the standard normal distribution, the areas are as follows:

z = 0 to 1.35: 0.411

z = 0 to 2.24: 0.487

z = 1.35 to 2.24: 0.076

Assuming that these refer to the standard normal distribution, the areas are as follows:

z = 0 to 1.35: 0.411

z = 0 to 2.24: 0.487

z = 1.35 to 2.24: 0.076

More answers

Assuming that these refer to the standard normal distribution, the areas are as follows:

z = 0 to 1.35: 0.411

z = 0 to 2.24: 0.487

z = 1.35 to 2.24: 0.076

Q: What is the area between z1.35 and z 2.24 and z0?

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The equation of a sphere with radius r, centered at (x0 ,y0 ,z0 ) is (x-x0 )+(y-y0 )+(z-z0 )=r2

The gradient, at any point P:(x, y, z), of a scalar point function Φ(x, y, z) is a vector that is normal to that level surface of Φ(x, y, z) that passes through point P. The magnitude of the gradient is equal to the rate of change of Φ (with respect to distance) in the direction of the normal to the level surface at point P. Grad Φ, evaluated at a point P:(x0, y0, z0), is normal to the level surface Φ(x, y, z) = c passing through point P. The constant c is given by c = Φ(x0, y0, z0).

lying on the same plane indicates that the normal vector to that plane will always be the same i.e to that point or vector. Another way would be that it should satisfy the standard equation of plane Ax+By+Cz = D in the case of a point or the dot product should equal zero. <A,B,C> . <x-x0, y-y0, z-z0> = 0

The Fundamental Theorem of Algebra:If P(z) = Σ­­nk=0 akzk where ak Є C, n ≥ 1, and an ≠ 0, then P(z0) = 0 for some z0 Є C. Descriptively, this says that any nonconstant polynomial over the complex number space, C, can be written as a product of linear factors.Proof:First off, we need to apply the Heine-Borel theorem to C. The Heine-Borel theorem states that if S is a closed and bounded set in an m-dimensional Euclidean space (written as Rm), then S is compact.From above, P(z) = Σ­­nk=0 akzk where ak Є C, n ≥ 1, and an ≠ 0. Let m = inf{|P(z)| : z Є C} where inf is the infinum, or the greatest lower bound of the set.From the triangle inequality, |P(reit)| ≥ rn(|an| - r-1|an-1| - … - r-n|a0|),so limr --> ∞ |P(reit)| = ∞. Therefore there is a real number R that |P(reit)| > m + 1 whenever r > R.If S = {reit : r ≤ R}, then S is compact in C, by the Heine-Borel Theorem; and let m = inf{|P(z)| : z Є S}. |P| is a continuous and real-valued function in S, so, using the result from another proof not done here, it has a minimum value on S; i.e., there is a value for z0 Є S that makes |P(z0)| = m. So, if m = 0 then the theorem is proved.We're going to show that m = 0 by proving that m can't equal anything else, and since we know m exists, it has no choice but to be zero. So, suppose m ≠ 0 and let Q(z) = P(z + z0)/P(z0), z Є C.Q is therefore a polynomial with degree n and |Q(z)| ≥ 1 for all z Є C.Q(0) = 1 so Q(z) can be expressed via P's series as:Q(z) = 1 + bkzk + … + bnzn where k is the smallest positive integer ≤ n such that bk ≠ 0.Since |-|bk|/bk| = 1, there exists a t0 Є [0, 2π/k] such that eikt0 = -|bk|/bk.Then Q(reit0) = 1 + bkrkeikt0 + bk+1rk+1ei(k+1)t0 + … + bnrneint0= 1 - rk|bk| + bk+1rk+1ei(k+1)t0 + … + bnrneint0.So, if rk|bk| < 1 then |Q(reit0)| ≤ 1 - rk(|bk| - r|bk+1| - … - rn-k|bn|).That means that if we pick a small enough r, we can make |Q(reit0)| ≤ 1 which contradicts the statement above that |Q(z)| ≥ 1 for all z Є C. Therefore m ≠ 0 doesn't hold and P(z0) = 0Q.E.D.Another proofSuppose P has no zeroes. Then we can define the function f(z) = 1 / P(z), and f is analytic. By the proof above, P(z) tends to infinity as z tends to infinity; hence f(z) tends to 0 as z tends to infinity. So there is a disc S such that f, restricted to the outside of S, is bounded. Also by the proof above, f is bounded inside the disc as well; therefore f is bounded. Now we apply a theorem called Liouville's Theorem, which says that any analytic function which is defined on all of C and is bounded must be a constant. So f is a constant; therefore P is constant. But we were assuming that P is not constant, so this is a contradiction. (To prove Liouville's Theorem: Suppose M is a bound for the function f, i.e. |f(z)| < M for all z. Suppose a and b are complex numbers, and we want to show f(a) = f(b). Use the theorem that f(a) = integral of f(z)/(z-a) / (2 * pi * i) around the circle of radius R and centre 0. Then, if R is sufficiently large:|f(b) - f(a)|= | integral, around circle, of (f(z) * (1/(z-b) - 1/(z-a))) | / (2*pi)= | integral around circle of (f(z) * (b-a) / ((z-a)(z-b)) ) | / (2*pi)

It's not pretty (because I've never written anything in C before but it worked nonetheless) but here you go. I compiled with Microsoft Visual C++ 2010 Express with a Win32 Console Application and this is everything that was written. It worked but looks kind of bulky. Tweak to your heart's content. I wasn't 100% sure what you wanted so I made three versions that do basically the same thing with slight alterations. I learned to code in C for you... This one lists each one that was either positive or negative. // snfp.cpp : Defines the entry point for the console application. // #include "stdafx.h" #include "stdio.h" int main (void) { /* declarations */ double x0=0, x1=0, x2=0, x3=0, x4=0, x5=0, x6=0, x7=0, x8=0, x9=0, y0=0, y1=0, y2=0, y3=0, y4=0, y5=0, y6=0, y7=0, y8=0, y9=0, z0=0, z1=0, z2=0, z3=0, z4=0, z5=0, z6=0, z7=0, z8=0, z9=0; /* executable statements */ printf ("Enter ten real numbers: "); scanf ("%lf %lf %lf %lf %lf %lf %lf %lf %lf %lf", &x0, &x1, &x2, &x3, &x4, &x5, &x6, &x7, &x8, &x9); if(x0<0){ y0 = x0;} if(x0>0){ z0 = x0;} if(x1<0){ y1 = x1;} if(x1>0){ z1 = x1;} if(x2<0){ y2 = x2;} if(x2>0){ z2 = x2;} if(x3<0){ y3 = x3;} if(x3>0){ z3 = x3;} if(x4<0){ y4 = x4;} if(x4>0){ z4 = x4;} if(x5<0){ y5 = x5;} if(x5>0){ z5 = x5;} if(x6<0){ y6 = x6;} if(x6>0){ z6 = x6;} if(x7<0){ y7 = x7;} if(x7>0){ z7 = x7;} if(x8<0){ y8 = x8;} if(x8>0){ z8 = x8;} if(x9<0){ y9 = x9;} if(x9>0){ z9 = x9;} printf ("\nThe negative numbers are %lf %lf %lf %lf %lf %lf %lf %lf %lf %lf.\n", y0, y1, y2, y3, y4, y5, y6, y7, y8, y9); printf ("\nThe positive number are %lf %lf %lf %lf %lf %lf %lf %lf %lf %lf.\n", z0, z1, z2, z3, z4, z5, z6, z7, z8, z9); printf ("Type something to exit"); scanf ("%lf", &x0); return (0); } This one tells you how many were positive and how many were negative. // snfp.cpp : Defines the entry point for the console application. // #include "stdafx.h" #include "stdio.h" int main (void) { /* declarations */ double x0=0, x1=0, x2=0, x3=0, x4=0, x5=0, x6=0, x7=0, x8=0, x9=0, y0=0, y1=0, y2=0, y3=0, y4=0, y5=0, y6=0, y7=0, y8=0, y9=0, z0=0, z1=0, z2=0, z3=0, z4=0, z5=0, z6=0, z7=0, z8=0, z9=0, neg=0, pos=0; /* executable statements */ printf ("Enter ten real numbers: "); scanf ("%lf %lf %lf %lf %lf %lf %lf %lf %lf %lf", &x0, &x1, &x2, &x3, &x4, &x5, &x6, &x7, &x8, &x9); if(x0<0){ y0 = 1;} if(x0>0){ z0 = 1;} if(x1<0){ y1 = 1;} if(x1>0){ z1 = 1;} if(x2<0){ y2 = 1;} if(x2>0){ z2 = 1;} if(x3<0){ y3 = 1;} if(x3>0){ z3 = 1;} if(x4<0){ y4 = 1;} if(x4>0){ z4 = 1;} if(x5<0){ y5 = 1;} if(x5>0){ z5 = 1;} if(x6<0){ y6 = 1;} if(x6>0){ z6 = 1;} if(x7<0){ y7 = 1;} if(x7>0){ z7 = 1;} if(x8<0){ y8 = 1;} if(x8>0){ z8 = 1;} if(x9<0){ y9 = 1;} if(x9>0){ z9 = 1;} neg = y0 + y1 + y2 + y3 + y4 + y5 + y6 + y7 + y8 + y9; pos = z0 + z1 + z2 + z3 + z4 + z5 + z6 + z7 + z8 + z9; printf ("\nThere are %lf negative numbers.\n", neg); printf ("\nThere are %lf positive numbers.\n", pos); printf ("Type something to exit"); scanf ("%lf", &x0); return (0); } This one sums the positive and then sums the negative. // snfp.cpp : Defines the entry point for the console application. // #include "stdafx.h" #include "stdio.h" int main (void) { /* declarations */ double x0=0, x1=0, x2=0, x3=0, x4=0, x5=0, x6=0, x7=0, x8=0, x9=0, y0=0, y1=0, y2=0, y3=0, y4=0, y5=0, y6=0, y7=0, y8=0, y9=0, z0=0, z1=0, z2=0, z3=0, z4=0, z5=0, z6=0, z7=0, z8=0, z9=0, neg=0, pos=0; /* executable statements */ printf ("Enter ten real numbers: "); scanf ("%lf %lf %lf %lf %lf %lf %lf %lf %lf %lf", &x0, &x1, &x2, &x3, &x4, &x5, &x6, &x7, &x8, &x9); if(x0<0){ y0 = x0;} if(x0>0){ z0 = x0;} if(x1<0){ y1 = x1;} if(x1>0){ z1 = x1;} if(x2<0){ y2 = x2;} if(x2>0){ z2 = x2;} if(x3<0){ y3 = x3;} if(x3>0){ z3 = x3;} if(x4<0){ y4 = x4;} if(x4>0){ z4 = x4;} if(x5<0){ y5 = x5;} if(x5>0){ z5 = x5;} if(x6<0){ y6 = x6;} if(x6>0){ z6 = x6;} if(x7<0){ y7 = x7;} if(x7>0){ z7 = x7;} if(x8<0){ y8 = x8;} if(x8>0){ z8 = x8;} if(x9<0){ y9 = x9;} if(x9>0){ z9 = x9;} neg = y0 + y1 + y2 + y3 + y4 + y5 + y6 + y7 + y8 + y9; pos = z0 + z1 + z2 + z3 + z4 + z5 + z6 + z7 + z8 + z9; printf ("\nThe negative numbers add up to %lf.\n", neg); printf ("\nThe positive number add up to %lf.\n", pos); printf ("Type something to exit"); scanf ("%lf", &x0); return (0); }

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