Write $P_k=P(X=k)$ for $k=0,1,2,\dots$, with ${\sum }_k{P}_k=1$. Provided the relevant moments exist, the expected value and variance areThe displayed support includes zero, so these are nonnegative integer values.

For $\mathbf{r}(t)=(t,t,t^2)$, the tangent vector is ${\mathbf{r}}^{\prime }(t)=(1,1,2t)$. Substitute into the field:Its scalar product with the tangent is $3t^2-4t-7t^6$. Hence the explicitly parametrized line integral isAs a check, the potential from part (d) has values $-128+C$ at $(2,2,4)$ and $-2+C$ at $(1,1,1)$, giving the same difference.

Evaluate the line integral along the three coordinate segments, treating coordinates not changing along a segment as constants. On the first segment, $y=z=0$, so $F_x=0$ and the contribution is zero. On the second segment,On the final segment,The fundamental theorem for line integrals therefore givesTaking its gradient verifies each component of the field. The constant is $C=\Phi (0,0,0)$; the field specifies only potential differences because the gradient of every constant is zero.

Differentiate the components directly:Thus $\boxed{\nabla \times \mathbf{F}=0}$. The field is smooth on all of ${\mathbb{R}}^3$, a simply connected domain, so the zero-curl condition is sufficient for a conservative vector field. The domain condition matters: zero curl alone need not imply a global potential on a region with holes.

For a twice continuously differentiable scalar potential, the curl of its gradient isEquality of mixed partial derivatives makes every component zero. ThusThis is the curl of a gradient identity.

Parametrize the path by $\mathbf{r}(t)=(x(t),y(t),z(t))$, from $t=a$ to $t=b$. For the conservative vector field $\mathbf{F}=\nabla \Phi$, the chain rule givesIntegrating proves the fundamental theorem for line integrals:Only the endpoints matter, so the integral is independent of the connecting path.

The sphere of centre $\mathbf{a}$ and radius $p$ has equationLet $d=|\mathbf{b}-\mathbf{a}|$ be the distance between the centres. A nondegenerate circle of intersection of two spheres exists exactly whenThe upper inequality excludes disjoint or externally tangent spheres. The lower excludes strict containment and internal tangency. It also excludes coincident centres, which give either no intersection or an entire common sphere rather than one circle. These inequalities are sufficient because a triangle with side lengths $p,q,d$ then exists with nonzero area, and rotating its third vertex about the line of centres traces a circle.

For a point on both spheres, expand their equations:Subtracting cancels ${\mathbf{r}}^2$ and gives the planeIts normal is parallel to the line of centres. Under the strict inequalities, the plane cuts either sphere in a genuine circle.

Expand the vector triple product in Cartesian components. Its first component iswhere adding $a_x{b}_x{c}_x$ to both products introduces cancelling terms. Likewise the other two components areTogether these proveNow use the cyclic property of the scalar triple product and the just-proved identity:This proves Lagrange's identity in the requested form,

By Bayes' theorem, provided $p+q>0$,If $p=q=0$, singing has probability zero and the stated observation cannot occur in this model; that conditional probability is undefined.

Let $S$ denote singing, and $M,F$ denote male and female. Equal population counts give $P(M)=P(F)=1/2$. The law of total probability therefore yields

Equating Cartesian components gives $1-s=t$, $-s=-1+t$, and $2+s=2t$. Substitution of $t=1-s$ into the last equation gives $s=0$, hence $t=1$. Both parametrizations therefore give the intersection point

The direction vectors are $v=(-1,-1,1)$ and $w=(1,1,2)$. Their scalar product is $v\cdot w=-1-1+2=0$. Both are nonzero, so the lines are perpendicular:

The initial condition gives $1=1/(C-1)$, hence $C=2$. ThusThe displayed interval is the maximal connected interval containing the initial point $x=1$; the solution of a differential equation diverges at $x=2$.

For $y\ne 0$, the separable differential equation gives $dy/y^2=dx$. Integrating gives $-1/y=x+C_0$, orEach solution is considered on an interval not containing its pole. The additional stationary solution $y=0$ is excluded by the stated condition.

Since $1+\cos 2x=2{\cos }^{2}x$, the equation reduces to ${\cos }^{2}x=0$. The only solution in the open interval $(\pi ,2\pi )$ is