How many records

Idea. Linearity of expectation turns the count into a sum of probabilities, and each probability follows from symmetry among the first $k$ values.

Step 1: indicators. The number of records is $R=I_1+\cdots +I_{10}$, where $I_k=1$ if $X_k$ is a record.

Step 2: one indicator. $X_k$ is a record exactly when it is the maximum of $X_1,\dots ,X_k$. The values are i.i.d. and continuous (no ties), so each of the $k$ values is the maximum with the same probability: $\mathbb{P}(I_k=1)=\frac{1}{k}$.

Step 3: add. $$\mathbb{E}[R]=1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{10}=\frac{7381}{2520}\approx \mathrm{2.929.}$$

Remarks. The answer does not depend on the distribution. With $n$ values there are about $\ln n$ records, so records become rare: in a million observations you expect only about $14$. The indicators are in fact independent, though linearity did not need it.