Daily math problems for October 11, 2026 — Descriptive Statistics & Sampling, Euclidean Geometry, Calculus & more

Row Y total $=20$. Grand total $=50$.

Fraction $={\displaystyle \frac{20}{50}}={\displaystyle \frac{2}{5}}$.

${180}^{\circ }-{110}^{\circ }={70}^{\circ }$.

Write $f(x)=x\cdot \sqrt{x^2+9}$ and apply the product rule, using the chain rule on the square root.

$$\frac{d}{dx}\sqrt{x^2+9}=\frac{x}{\sqrt{x^2+9}}$$

$$f^{\mathrm{\prime }}(x)=\sqrt{x^2+9}+x\cdot \frac{x}{\sqrt{x^2+9}}=\sqrt{x^2+9}+\frac{x^2}{\sqrt{x^2+9}}$$

At $x=4$, $\sqrt{x^2+9}=\sqrt{25}=5$.

$$f^{\mathrm{\prime }}(4)=5+\frac{16}{5}=\frac{41}{5}$$

Standardize the two cut-offs: $$z_{63}=\frac{63-58}{5}=1,z_{53}=\frac{53-58}{5}=-1$$ $$P(X>63)=1-\mathrm{\Phi }(1)=0.1587$$ $$P(X>53)=\mathrm{\Phi }(1)=0.8413$$ Since being large guarantees being medium, the conditional probability is $$P(\text{large}\mid \text{medium})=\frac{0.1587}{0.8413}=0.189$$

The general term is $$T_k=\left(\genfrac{}{}{0px}{}{5}{k}\right)(2x{)}^{5-k}(3{)}^k\mathrm{.}$$ For the $x^3$ term, set the exponent of $x$ equal to $3$: $$5-k=3⟹k=2.$$ Substitute $k=2$: $$\left(\genfrac{}{}{0px}{}{5}{2}\right)(2x{)}^3(3{)}^2=10\cdot 8x^3\cdot 9.$$ So the coefficient is $$10\cdot 8\cdot 9=720.$$

$m_1\cdot m_2=-1\Rightarrow m_2=-{\displaystyle \frac{1}{2}}$.

$5={\displaystyle \frac{(x-2)+(x+4)}{2}}\Rightarrow 10=2x+2\Rightarrow x=4$.

Using the closed forms $\cos {36}^{\circ }=\frac{1+\sqrt{5}}{4}$ and $\cos {72}^{\circ }=\frac{\sqrt{5}-1}{4}$, the difference is $\frac{(1+\sqrt{5})-(\sqrt{5}-1)}{4}=\frac{2}{4}=\frac{1}{2}$.

Let $u=x$ and ${\displaystyle \frac{\mathrm{d}v}{\mathrm{d}x}}=e^{2x}$, so ${\displaystyle \frac{\mathrm{d}u}{\mathrm{d}x}}=1$ and $v=\frac{1}{2}{e}^{2x}$.

By parts, ${\displaystyle \int u\mathrm{d}v=uv-\int v\mathrm{d}u}$: $$\int x{e}^{2x}\mathrm{d}x=\frac{1}{2}x{e}^{2x}-\int \frac{1}{2}{e}^{2x}\mathrm{d}x\mathrm{.}$$

The remaining integral gives $\frac{1}{4}{e}^{2x}$, so the antiderivative is $$\frac{1}{2}x{e}^{2x}-\frac{1}{4}{e}^{2x}\mathrm{.}$$

Evaluate from $0$ to $1$: $$(\frac{1}{2}{e}^2-\frac{1}{4}{e}^2)-(0-\frac{1}{4})=\frac{1}{4}{e}^2+\frac{1}{4}\mathrm{.}$$

Hence the value is $$\frac{e^2+1}{4}\mathrm{.}$$

Factor the $n$-th term to expose the first term and the ratio: $$\frac{3^{n+1}}{4^{n-1}}=9{\left(\frac{3}{4}\right)}^{n-1}$$

So the first term is $a=9$ (the $n=1$ value) and the common ratio is $$r=\frac{3}{4}$$

Since $\mathrm{\mid }r\mathrm{\mid }=\frac{3}{4}<1$, the series converges to $$S=\frac{a}{1-r}=\frac{9}{1-\frac{3}{4}}=\frac{9}{\frac{1}{4}}=36$$