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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 $n=2$.
$P(X=1)=\binom{2}{1}\cdot\dfrac{1}{13}\cdot\dfrac{12}{13}=\dfrac{24}{169}$
$P(X=2)=\binom{2}{2}\cdot\left(\dfrac{1}{13}\right)^2=\dfrac{1}{169}$
$P(X=1)+P(X=2)=\dfrac{24}{169}+\dfrac{1}{169}=\dfrac{25}{169}$
Sharks belong to which class of vertebrates?
Class Chondrichthyes includes cartilaginous fish (sharks, rays). Osteichthyes are bony fish. Echinodermata are starfish etc.; Urochordata are tunicates.
Compare the quantities:
Column A: $\sqrt{x^4 + 6x^2 + 9}$
Column B: $x^2 + 3$
Notice that the expression under the square root is a perfect square:
$x^4 + 6x^2 + 9 = (x^2)^2 + 2(x^2)(3) + 3^2 = (x^2 + 3)^2$
Therefore: $\sqrt{x^4 + 6x^2 + 9} = \sqrt{(x^2 + 3)^2} = |x^2 + 3|$
Since $x^2 \geq 0$ for all real $x$, we have $x^2 + 3 \geq 3 > 0$, which means $x^2 + 3$ is always positive.
Thus: $|x^2 + 3| = x^2 + 3$
Therefore, Column A equals Column B, so the answer is C (the two quantities are equal).
Gestational Diabetes Mellitus (GDM) — Diagnosis by OGTT (WHO 2013 / IADPSG criteria):
| Test | Threshold | This Patient |
|---|---|---|
| Fasting | \(\geq 5.1\,\text{mmol/L}\) | \(5.8\,\text{mmol/L}\) ✓ |
| 1-hour | \(\geq 10.0\,\text{mmol/L}\) | Not tested |
| 2-hour | \(\geq 8.5\,\text{mmol/L}\) | \(9.2\,\text{mmol/L}\) ✓ |
Initial Management (NICE/ADA guidelines):
- Medical Nutrition Therapy (MNT): Carbohydrate control (40–45% of calories), low GI foods, 3 meals + 2–3 snacks/day
- Physical activity: 30 min moderate exercise daily (e.g., walking)
- Self-monitoring blood glucose (SMBG) 4× daily
- Reassess in 1–2 weeks
Target glucose levels:
- Fasting: \(\leq 5.3\,\text{mmol/L}\)
- 1-hour post-meal: \(\leq 7.8\,\text{mmol/L}\)
- 2-hour post-meal: \(\leq 6.7\,\text{mmol/L}\)
If targets not met with MNT after 2 weeks → Add insulin (preferred) or metformin. Insulin does not cross placenta; metformin does but is used in many guidelines.
In 1950, if the printing cost per newspaper was $0.05, what would have been the total cost of printing the average daily circulation?
[Note: From the graph, average daily circulation in 1950 was 520,000 newspapers]
Total cost = (number of newspapers) × (cost per newspaper)
Total cost $= 520{,}000 \times 0.05$
$= 520{,}000 \times \frac{5}{100} = 520{,}000 \times \frac{1}{20} = \frac{520{,}000}{20} = 26{,}000$
Therefore, the total cost would be $26,000.
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