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tino4ka555 [31]
3 years ago
9

You are to give an injection of a drug. The dosage is 0.4 mg per kilogram of bod The concentration of the drug in vial is listed

as 500 ug/ml. The patient's chart Hists weight as 168 pounds. How many milliliters (= cc) are you to inject? Patient's weight Concentration of drug Show calculations: mg/ml
Mathematics
1 answer:
shusha [124]3 years ago
3 0

Answer:

We inject 60.9628 milliliters of dosage.

Step-by-step explanation:

Given : You are to give an injection of a drug. The dosage is 0.4 mg per kilogram of bod The concentration of the drug in vial is listed as 500 ug/ml. The patient's chart Hosts weight as 168 pounds.

To find : How many milliliters (= cc) are you to inject?

Solution :

We know that,

1 pound = 0.453592 kg,

Patient's weight = 168  pounds

Patient's weight in kg =168\times 0.453592=76.203456  kg

The dosage is 0.4 mg per kilogram of bod.

So, dosage of patient weight = 0.4 \times 76.2035\ mg = 30.4814\ mg

We know,  1 micro gram = 0.001 mg

Concentration of drug is given by,

500\ \text{micro grams/ml}= 500\times 0.001\ \text{mg/ml} = 0.5\ \text{mg/ml}

Now we are supposed to find milliliters inject,

So, milliliters of dosage required to inject is

d= \frac{30.4814}{0.5} = 60.9628

Therefore, We inject 60.9628 milliliters of dosage.

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Construct a 99% confidence interval for the population standard deviation of white blood cell count (in cells per microliter). A
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The correct option is 1.148 < σ < 6.015

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a) 0.9959 = 99.59% probability that the mean weight is less than 9.3 ounces

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Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

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The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

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Mean of 8.0 ounces and a standard deviation of 1.1 ounces.

This means that \mu = 8, \sigma = 1.1

(a) If 5 potatoes are randomly selected, find the probability that the mean weight is less than 9.3 ounces?

n = 5 means that s = \frac{1.1}{\sqrt{5}} = 0.4919

This probability is the pvalue of Z when X = 9.3. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{9.3 - 8}{0.4919}

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Z = \frac{X - \mu}{s}

Z = \frac{9 - 8}{0.4491}

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Z = 2.23 has a pvalue of 0.9871

1 - 0.9871 = 0.0129

0.0129 = 1.29% probability that the mean weight is more than 9.0 ounces

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