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JulsSmile
10 days ago
11

You’re working in a pharmacy, and are making a table to help with dosage amounts for a certain drug. The recommended dosage is 4

0 milligrams per 2.2 pounds of body weight, divided into three daily doses and taken every 8 hours. You want entries for children weighing 10 pounds, 15 pounds and 30 pounds. The drug comes in a 200 mg5 ml formula. If 1 teaspoon = 5 ml, how many teaspoons are required for each entry? Round your answer to the nearest 14 teaspoon.
Mathematics
1 answer:
lawyer [4K]10 days ago
7 0

Answer: Children who weigh 10 pounds require 12 teaspoons, those at 15 pounds need 18 teaspoons, and for 30 pounds, it is necessary to have 36 teaspoons.

Step-by-step explanation:

We have a recommended dose of 40 mg per 2.2 pounds of weight, given in three doses throughout the day every 8 hours.

Thus, for a single pound, the daily required dosage is calculated as \dfrac{40}{2.2}=18.18\ ml.

Each individual dose corresponds to \dfrac{18.18}{3}=6.06\ ml of the drug.

This means for one pound, the dosage needed per dose is 6.06 ml.

Thus, for 10 pounds, the calculation is 10 x 6.06 ml = 60.6 ml

To convert ml to teaspoons, knowing that 1 teaspoon equals 5 ml, we find: (60.6 ml ÷ 5) = 12.12 ≈ 12 teaspoons.

For a weight of 15 pounds, the drug requirement is 15 x 6.06 ml = 90.9 ml.

This translates to (90.9 ml ÷ 5) = 18.18 ≈ 18 teaspoons.

For the weight of 30 pounds, it is calculated as 30 x 6.06 ml = 181.8 ml.

Converting this gives us (181.8 ml ÷ 5) = 36.36 ≈ 36 teaspoons.

Therefore, children weighing 10 pounds need 12 teaspoons, those at 15 pounds require 18 teaspoons, and those at 30 pounds need 36 teaspoons.

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Justin wants to buy plants for his property, and he can afford to spend at most $480. The local nursery sells holly bushes for $
lawyer [4039]

Answer: 10 holly bushes and 12 bayberry shrubs

Step-by-step explanation: use your brain or a calculator to figure out the cost of each type of plant and then total them up

3 0
14 days ago
Suppose that the number of drivers who travel between a particular origin and destination during a designated time period has a
zzz [4035]

Answer:

Step-by-step explanation:

It has been established that the count of drivers traveling between a specific origin and destination in a certain time frame follows a Poisson distribution with a mean μ = 20 (as indicated in the article "Dynamic Ride Sharing: Theory and Practice"†).

a) P(X\leq 15) = 0.1565=0.157

b) P(X>26) =1-F(26)\\= 1-0.9221\\=0.0779=0.078

c) P(15\leq x\leq 26)\\=F(26)-F(14)\\=0.9221-0.1049\\=0.8172=0.817

d) 2 standard deviations = 2(20) = 40

Thus, this means the range for 2 standard deviations is

20-40, 20+40

which equates to (0,60)

P(0

5 0
2 days ago
Sin 1 + sin 2 + sin3 .................+ sin 360 = ?
PIT_PIT [3949]

Assuming arcs are measured in degrees, let S represent the following sum:

S = sin 1° + sin 2° + sin 3° +... + sin 359° + sin 360°


By rearranging, S can be reformulated as

S = [sin 1° + sin 359°] + [sin 2° + sin 358°] +... + [sin 179° + sin 181°] + sin 180° +
+ sin 360°

S = [sin 1° + sin(360° – 1°)] + [sin 2° + sin(360° – 2°)] +... + [sin 179° + sin(360° – 179)°]
+ sin 180° + sin 360°          (i)


However, for any real k,

sin(360° – k) = – sin k


Thus,

S = [sin 1° – sin 1°] + [sin 2° – sin 2°] +... + [sin 179° – sin 179°] + sin 180° + sin 360°

S results in 0 + 0 +... + 0 + 0 + 0        (... since sine of 180° and 360° are both equal to 0)

Therefore, S equals 0.


Each pair within the brackets negates itself, leading the sum to total zero.

∴   sin 1° + sin 2° + sin 3° +... + sin 359° + sin 360° equals 0.          ✔


I hope this clarifies things. =)


Tags:  sum summatory trigonometric trig function sine sin trigonometry

4 0
4 days ago
A random sample of 16 students selected from the student body of a large university had an average age of 25 years and a standar
PIT_PIT [3949]

Answer:

The P-value ranges between 2.5% and 5% according to the t-table.

Step-by-step explanation:

A random sample of 16 students from a large university showed an average age of 25 years with a standard deviation of 2 years.

Let \mu = true average age of all students at the university.

So, the Null Hypothesis, H_0 : \mu \leq 24 years {indicating the average age is less than or equal to 24 years}

Alternate Hypothesis, H_A : \mu > 24 years {indicating the average age is significantly greater than 24 years}

Here we employ the One-sample t-test statistics as the population's standard deviation is unknown;

                              T.S. = \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

where, \bar X = sample average age = 25 years

             s = sample standard deviation = 2 years

             n = sample size = 16

This gives us the test statistics = \frac{25-24}{\frac{2}{\sqrt{16} } } ~ t_1_5

                                     = 2

The value of the t-test statistics is 2.

Moreover, the P-value of the test-statistics can be found as follows;

P-value = P(t_1_5 > 2) = 0.034 {as per the t-table}

Thus, the P-value lies between 2.5% and 5% based on the t-table.

8 0
15 days ago
Suppose the coordinate of a is 0, ar = 5, and at =7. What are the possible coordinates of the midpoint of rt ?
Zina [3917]

Here, 'a' relates to 0.

There are two scenarios for 'r' and 't'.

Scenario 1.

Both are positioned on the same side to the right of 'a'.

In this case, 'r' would equal 5, and 't' would equal 7.

The midpoint between 'r' and 't' is \frac{5+7}{2} =6.

Scenario 2.

If both are found to the left of 'a'.

Then 'r' would equal -5, while 't' would equal -7.

The midpoint is \frac{-7-5}{2} =-6.

Scenario 3.

If 'r' is right of 'a' and 't' is left of 'a'.

Thus 'r' equals 5 and 't' equals -7.

The midpoint is \frac{-7+5}{2}=-1.

Scenario 4.

If 'r' is left of 'a' while 't' is right of 'a'.

In this case, 'r' corresponds to -5 and 't' corresponds to 7.

The midpoint is \frac{-5+7}{2}=1.

The potential midpoint coordinates for 'rt' are 6, -6, 1, and -1.

4 0
1 day ago
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