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If an intravenous solution containing 123 mg of a drug substance in each 250-mL bottle is to be administered at the rate of 200
1 answer:
Answer:
24.39mL of the solution would be given per hour.
Step-by-step explanation:
This problem can be solved by direct rule of three, in which there are a direct relationship between the measures, which means that the rule of three is a cross multiplication.
The first step to solve this problem is to see how many mg of the solution is administered per hour.
Each minute, 200 ug are administered. 1mg has 1000ug, so
1mg - 1000 ug
xmg - 200 ug
In each minute, 0.2 mg are administered. Each hour has 60 minutes. How many mg are administered in 60 minutes?
1 minute - 0.2 mg
60 minutes - x mg
In an hour, 12 mg of the drug is administered. In 250 mL, there is 123 mg of the drug. How many ml are there in 12 mg of the drug.
123mg - 250mL
12 mg - xmL
mL
24.39mL of the solution would be given per hour.
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Answer:
∠B ≅ ∠Y △ABC ~ △ZYX by the SAS similarity theorem.
Step-by-step explanation:
1. units
units
units
units, then
2. and
are right angles - given
3. two right angles are always congruent.
4. by SAS similarity theorem.
SAS Similarity Theorem states that if two sides in one triangle are proportional to two sides in another triangle and the included angle in both are congruent, then the two triangles are similar.
Answer:
a) There is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.
b) There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.
c) There is a 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation
.
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
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.
In this problem, we have that:
The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3 minutes. This means that .
(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?
We are working with a sample mean of 37 jets. So we have that:
Total time of 320 minutes for 37 jets, so
This probability is the pvalue of Z when . So
has a pvalue of 0.7422. This means that there is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.
(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes?
Total time of 275 minutes for 37 jets, so
This probability is subtracted by the pvalue of Z when
has a pvalue of 0.0548.
There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.
(c) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes?
Total time of 320 minutes for 37 jets, so
Total time of 275 minutes for 37 jets, so
This probability is the pvalue of Z when subtracted by the pvalue of Z when
.
So:
From a), we have that for , we have
, that has a pvalue of 0.7422.
From b), we have that for , we have
, that has a pvalue of 0.0548.
So there is a 0.7422 - 0.0548 = 0.6874 = 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.