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Makovka662 [10]
3 years ago
15

Just as you are recovering, your child falls ill. in your medication dosages, so you hurriedly scan the Internet to make sure th

at what the doctor is giving your child is safe. The doctor writes the order for the following: You know that there have been mistakes 150 mg of an antibiotic to be given every 6 hours Your child weighs 30 lb. Your research indicates that 20-40 mg/kg/day is the recommended dosage. Is this prescription safe for your child? Why or why not?
Mathematics
1 answer:
Bumek [7]3 years ago
5 0

Answer:

Given dose (i.e 600 mg) lies outside the recommended range of  271.8-543.6 mg/day that too on the higher side

Hence, the prescription is not safe.

Step-by-step explanation:

Given:

Weight of medicine given per interval = 150 mg

time interval = 6 hours

thus, number of intervals per day = \frac{\textup{24}}{\textup{6}} = 4

therefore,

the total dose of medicine provided per day = 4 × 150 = 600 mg

Now,

Recommended dosage =  20-40 mg/kg/day

weight of child = 30 lb

also,

1 lb = 0.453 kg

thus,

weight of child = 30 × 0.453 = 13.59 kg

Therefore, the recommended dose for the child

=  (  20-40 mg/kg/day ) × 13.59

= 271.8-543.6 mg/day

now,

the given dose (i.e 600 mg) lies outside the recommended range of  271.8-543.6 mg/day that too on the higher side

Hence, the prescription is not safe.

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AleksandrR [38]
Let's start by assuming Armando's house is between Joey's and the park. 

Let x be the distance Joey walked to Armando's house.

<span>The park is 9/10 mile from Joey's home. Joey leaves home and walks to Armando's home. Then Joey and Armando walk 3/5 mile to the park. 

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x = \dfrac{9}{10} - \dfrac{3}{5} = \dfrac{9}{10} -\dfrac{6}{10} = \dfrac{3}{10}

That's probably the answer they're looking for.  But what if the park is between Joey and Armando's houses or Joey is between the park and Armando?  (The latter isn't really possible with the given distances.)

Let a, b, c be the distances between three collinear points like we have here.  Our equation is really a few equations in one, something like

\pm a \pm b = \pm c

Let's get rid of the plus/minuses. Squaring,

a^2 + b^2\pm 2ab = c^2

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For us, that's a quadratic equation for c^2

4(9/10)^2(3/5)^2= (c^2-(9/10)^2 - (3/5)^2)^2

I'll skip right to the solutions,

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We could have gotten the 3/2 just by adding 9/10+3/5 but this was more fun.

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The number of students that are on the track team are 18.

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