The expectation value of an operator $A$ for a quantum state $|\psi\rangle$ is given by $\langle A \rangle = \langle \psi | A | \psi \rangle$. The given state is $|\psi\rangle = 0.8|\uparrow\rangle + 0.6|\downarrow\rangle$. The operator is $A = 10\sigma_z + 5\sigma_x$. We need to calculate $\langle A \rangle$. Since the expectation value is linear, $\langle A \rangle = 10\langle \sigma_z \rangle + 5\langle \sigma_x \rangle$. The basis states $|\uparrow\rangle$ and $|\downarrow\rangle$ are eigenstates of $\sigma_z$ with eigenvalues $+1$ and $-1$, respectively. In this basis, the Pauli matrices are:

$\sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$

$\sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$

The state vector in this basis is $|\psi\rangle = \begin{pmatrix} 0.8 \\ 0.6 \end{pmatrix}$. The corresponding bra vector is $\langle\psi| = \begin{pmatrix} 0.8 & 0.6 \end{pmatrix}$.

Calculate $\sigma_z$ Expectation Value

The expectation value of $\sigma_z$ is:

$\langle \sigma_z \rangle = \langle\psi|\sigma_z|\psi\rangle = \begin{pmatrix} 0.8 & 0.6 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 0.8 \\ 0.6 \end{pmatrix}$

$= \begin{pmatrix} 0.8 & -0.6 \end{pmatrix} \begin{pmatrix} 0.8 \\ 0.6 \end{pmatrix}$

$= (0.8)(0.8) + (-0.6)(0.6) = 0.64 - 0.36 = 0.28$

Calculate $\sigma_x$ Expectation Value

The expectation value of $\sigma_x$ is:

$\langle \sigma_x \rangle = \langle\psi|\sigma_x|\psi\rangle = \begin{pmatrix} 0.8 & 0.6 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0.8 \\ 0.6 \end{pmatrix}$

$= \begin{pmatrix} 0.6 & 0.8 \end{pmatrix} \begin{pmatrix} 0.8 \\ 0.6 \end{pmatrix}$

$= (0.6)(0.8) + (0.8)(0.6) = 0.48 + 0.48 = 0.96$

Combined Expectation Value Calculation

Now, we combine the expectation values:

$\langle 10\sigma_z + 5\sigma_x \rangle = 10\langle \sigma_z \rangle + 5\langle \sigma_x \rangle$

$= 10(0.28) + 5(0.96)$

$= 2.8 + 4.8 = 7.6$

The expectation value is $7.6$.