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Find the circle passing through the foci of the ellipse $\dfrac{x^2}{16}+\dfrac{y^2}{9}=1$ and having its centre at $(0,3)$.
For the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$: $a^2=16, b^2=9$, so $c^2=16-9=7$, giving foci at $(\pm\sqrt{7},0)$.
The circle has centre $(0,3)$. Radius $r=$ distance from $(0,3)$ to $(\sqrt{7},0)$:
$r^2=(\sqrt{7}-0)^2+(0-3)^2=7+9=16\Rightarrow r=4$
Equation: $x^2+(y-3)^2=16\Rightarrow x^2+y^2-6y+9=16\Rightarrow x^2+y^2-6y-7=0$
Evaluate $\displaystyle\lim_{n\to\infty}\left(\dfrac{(n+1)(n+2)\cdots3n}{n^{2n}}\right)^{1/n}$
$\ln L = \lim_{n\to\infty}\dfrac{1}{n}\sum_{r=1}^{2n}\ln\!\left(1+\dfrac{r}{n}\right) = \int_0^2\ln(1+x)\,dx$
$=\left[(1+x)\ln(1+x)-(1+x)\right]_0^2 = (3\ln3-3)-(-1) = 3\ln3-2$
$L = e^{3\ln3-2} = \dfrac{27}{e^2}$
Two cards are drawn one after another with replacement from a well-shuffled standard deck of 52 cards. Let $X$ be the number of aces obtained. Find $P(X=1)+P(X=2)$.
$p=P(\text{ace})=\dfrac{4}{52}=\dfrac{1}{13}$, $q=\dfrac{12}{13}$. Binomial with $n=2$.
$P(X=1)=\binom{2}{1}\cdot\dfrac{1}{13}\cdot\dfrac{12}{13}=\dfrac{24}{169}$
$P(X=2)=\binom{2}{2}\cdot\left(\dfrac{1}{13}\right)^2=\dfrac{1}{169}$
$P(X=1)+P(X=2)=\dfrac{24}{169}+\dfrac{1}{169}=\dfrac{25}{169}$
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