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A test paper has 5 questions, each with three answer choices of which only one is correct. What is the probability that a student scoring purely by guessing gets 4 or more questions right?
This is a binomial distribution with $n=5$, $p=\dfrac{1}{3}$ (correct), $q=\dfrac{2}{3}$ (wrong).
$P(X \geq 4) = P(X=4) + P(X=5)$
$P(X=4) = \binom{5}{4}\left(\dfrac{1}{3}\right)^4\left(\dfrac{2}{3}\right)^1 = 5 \cdot \dfrac{1}{81} \cdot \dfrac{2}{3} = \dfrac{10}{243}$
$P(X=5) = \binom{5}{5}\left(\dfrac{1}{3}\right)^5 = \dfrac{1}{243}$
$P(X \geq 4) = \dfrac{10}{243} + \dfrac{1}{243} = \dfrac{11}{243} = \dfrac{11}{3^5}$
Sudden Infant Death Syndrome (SIDS) β sudden, unexplained death of an infant under 1 year during sleep, remaining unexplained after thorough investigation.
Peak age: 2β4 months. Rare after 6 months.
Evidence-based SIDS Prevention (AAP Safe Sleep Guidelines 2022):
| Recommendation | Evidence Level |
|---|---|
| Supine (back) sleeping position β always | Strongest evidence β reduces SIDS by ~50% |
| Firm, flat sleep surface (crib/bassinet) | Strong β no soft mattresses, pillows, bumpers |
| Room-sharing (parent's room) WITHOUT bed-sharing | Strong β bed-sharing increases risk |
| Avoid smoke exposure (prenatal + postnatal) | Strong β smoking is major modifiable risk factor |
| Avoid overheating | Moderate β keep room 20β22Β°C |
| Breastfeeding | Moderate β protective effect |
| Pacifier use at sleep (after BF established) | Moderate β protective |
Prone sleeping increases SIDS risk 3β9 fold. The 'Back to Sleep' campaign reduced SIDS deaths by >50% since 1992.
Let $x$ and $m$ be positive numbers, where $m$ is a multiple of 3. Compare:
Quantity A: $\dfrac{x^m}{x^3}$
Quantity B: $x^{m/3}$
Simplify Quantity A: $\dfrac{x^m}{x^3} = x^{m-3}$.
Quantity B: $x^{m/3}$.
We need to compare $x^{m-3}$ with $x^{m/3}$. The relationship depends on the base $x$ and the exponent comparison.
Case 1: $x = 1$. Both quantities $= 1$. Equal.
Case 2: $x = 2,\ m = 6$. Qty A $= 2^{6-3} = 2^3 = 8$. Qty B $= 2^{6/3} = 2^2 = 4$. A > B.
Case 3: $x = 2,\ m = 3$. Qty A $= 2^{3-3} = 2^0 = 1$. Qty B $= 2^{3/3} = 2^1 = 2$. B > A.
Because different values give different results, the relationship cannot be determined.
Substituting the values:
$r = \frac{\sqrt{3} \times 4.29}{4} \approx \frac{1.732 \times 4.29}{4} \approx 1.857\text{ \AA}$. Thus, the radius is approximately $1.86\text{ \AA}$.
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