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an antibiotic is available as a solution with a concentration of 175 mg…

Question

an antibiotic is available as a solution with a concentration of 175 mg per 5 ml of solution or in 175 mg capsules. the recommended dosage of the antibiotic is 35 mg per day per kg. complete parts (a) and (b).
a. how many capsules should a 40 kg person take every 6 hours?
a 40 kg person should take \\( \square \\) capsule(s) every 6 hours.
(round to the nearest whole number as needed.)
b. suppose the antibiotic solution is given intravenously to a 40 kg person over a 6 hour period, using a metering system with a drop factor of 40 (drops/ml). what is the drop rate that should be used, in units of (drops/h)?
the drop rate should be \\( \square \\) (drops/h).
(round to the nearest whole number as needed.)

Explanation:

Part (a)

Step 1: Find daily dosage for 40 kg person

The recommended dosage is \( 35 \frac{\text{mg}}{\text{kg per day}} \). For a 40 kg person, daily dosage \( = 35 \times 40 = 1400 \) mg/day.

Step 2: Dosage every 6 hours

There are \( \frac{24}{6}=4 \) intervals in a day. So dosage per 6 hours \(=\frac{1400}{4} = 350 \) mg.

Step 3: Number of capsules per 6 hours

Each capsule has 175 mg. Number of capsules \(=\frac{350}{175}=2\).

Step 1: Find total volume for daily dosage

First, find total mg per day for 40 kg: \( 35\times40 = 1400 \) mg/day. Each 5 mL solution has 175 mg. Let volume be \( V \). Then \( \frac{175}{5}=\frac{1400}{V} \), so \( V=\frac{1400\times5}{175}=40 \) mL/day.

Step 2: Volume over 6 hours

In 6 hours (a quarter of a day), volume \(=\frac{40}{4}=10 \) mL.

Step 3: Calculate drop rate

Drop factor is \( 40 \frac{\text{drops}}{\text{mL}} \). Drops in 6 hours \(=10\times40 = 400 \) drops. Time is 6 hours. Drop rate \(=\frac{400}{6}\approx67 \) (rounded to nearest integer) \( \frac{\text{drops}}{\text{h}} \). Wait, wait, let's re - do with correct steps:

Alternative Step 1: Dosage per 6 hours is 350 mg (from part a). Each 5 mL has 175 mg, so volume for 350 mg: \( \frac{350}{175}\times5 = 10 \) mL (in 6 hours).

Step 2: Drop factor \( 40 \frac{\text{drops}}{\text{mL}} \), so drops in 6 hours: \( 10\times40=400 \) drops.

Step 3: Drop rate (drops per hour) \(=\frac{400}{6}\approx67 \) (or more accurately, let's use the formula:

Drop rate \(=\frac{\text{Volume (mL)}\times\text{Drop factor (drops/mL)}}{\text{Time (h)}} \)

Volume in 6 hours is 10 mL, time is 6 h, drop factor 40 drops/mL.

Drop rate \(=\frac{10\times40}{6}=\frac{400}{6}\approx67 \) (rounded to nearest integer). Wait, but let's check the daily volume again. Wait, the problem says "over a 6 - hour period". Wait, maybe I made a mistake earlier. Let's re - calculate:

From part (a), dosage per 6 hours is 350 mg. Each capsule is 175 mg, so 2 capsules (350 mg) which is in \( \frac{350}{175}\times5 = 10 \) mL (since 175 mg in 5 mL).

Drop factor is 40 drops/mL. So number of drops in 10 mL: \( 10\times40 = 400 \) drops. Time is 6 hours.

Drop rate \(=\frac{400}{6}\approx67 \) drops per hour. But let's do it with the formula for IV drop rate:

\( \text{Drop rate}=\frac{\text{Volume (mL)}\times\text{Drop factor (drops/mL)}}{\text{Time (min)}}\times60 \)? No, wait, the question asks for drops per hour.

Time is 6 hours = 360 minutes. But actually, the standard formula for drop rate (drops per minute) is \( \frac{\text{Volume (mL)}\times\text{Drop factor (drops/mL)}}{\text{Time (min)}} \), but here we need drops per hour.

Volume in 6 hours: 10 mL. Drop factor 40 drops/mL, so drops = 10*40 = 400 drops. Time = 6 h. So drop rate \(=\frac{400}{6}\approx67 \) drops per hour.

Wait, another way:

Daily dosage for 40 kg: 35*40 = 1400 mg.

Concentration: 175 mg per 5 mL, so concentration \( C=\frac{175}{5}=35 \) mg/mL.

Volume per day: \( V=\frac{1400}{35}=40 \) mL/day.

Volume per 6 hours: \( \frac{40}{4}=10 \) mL (since 24/6 = 4).

Drop factor: 40 drops/mL.

Drops in 6 hours: 10*40 = 400 drops.

Drop rate (drops per hour): \( \frac{400}{6}\approx67 \) (rounded to nearest integer).

Answer:

2

Part (b)