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Answer:
The doctor's dosage was not appropriate.
The right dosage is from 56.625 mg to 113.25 mg of Antibiotic to be given every 6 hour for the child with weight 25 lbs.
Step-by-step explanation:
It is given that 20 to 40 mg/kg/day is the recommended dosage.
Now, the doctor's order is 150 mg of antibiotic to be given every 6 hours. And my child weighs 25 lbs.
Now, 1 lb is equivalent to 0.453 kg.
So, my child's weight is (25 × 0.453) = 11.325 kg.
So, the doctor's order is 150 mg of antibiotic to be given every 6 hours for a child of weight 11.325 kg.
Hence, the dosage is [(150 × 4) ÷ 11.325] = 52.98 mg/kg/day.
So, this is not within the limit of 20 to 40 mg/kg/day.
Therefore, the doctor's dosage was not appropriate.
Now, let the appropriate dosage is x mg per every 6 hours for a child with weight 25 lbs i.e. 11.325 kg.
So,
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So, the right dosage is from 56.625 mg to 113.25 mg of antibiotic to be given every 6 hours for the child with weight 25 lbs. (Answer)
The given exponential function can be used to determine the
population of the fish at different times.
Response:
1. 200
2. 0.05
3. Please find attached the graph of the exponential function and the table
4. Range; [0, ∞)
Domain: [200, ∞)
5. 346 fishes (approximate value)
<h3>How is the population of the fish obtained from the exponential function?</h3>The given function is f(x) = 200·(1.05)ˣ
1. The initial population is given by the value of the function when x = 0,
which is given as follows;
At x = 0, f(0) = 200·(1.05)⁰ = 200
- The initial population of the fish is <u>200</u>
2. The rate of growth of the fish from the exponential function is given as follows;
The growth factor = 1.05
- The growth rate =<u> 0.05</u>
3. Please find attached the graph of the exponential function created
with MS Excel.
The asymptote of the exponential function is the line, <u>y = 0</u>
The table of values is presented as follows;
4. The domain is the possible x-values which is; 0 ≤ x < ∞
The range is the possible y-values which is; 200 ≤ f(x) < ∞
Which gives;
- Domain; [0, ∞)
- Range: [200, ∞)
5. The population of the fish after 11 years is f(11) = 200·(1.05)¹¹
f(11) = 200 × (1.05)¹¹ ≈ 342
- The population of the fish after 11 years is <u>346 (fishes)</u>
Learn more about exponential function here:
