A lateral thoracic spine on a 33-cm patient is usually taken using 100 mA (large focal spot), 0.5 seconds, 86 kVp, 40-inch SID,
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Answer:
a) 13.68% probability that if engineers inspect 20 buildings they will find exactly one that is structurally compromised.
b) 17.56% probability that if engineers inspect 20 buildings they will find less than 2 that are structurally compromised
c) 17.02% probability that if engineers inspect 20 buildings they will find greater than 4 that are structurally compromised
d) 75.70% probability that if engineers inspect 20 buildings they will find between 2 and 5 (inclusive) that are structurally compromised
e) The expected number of buildings that an engineer will find structurally compromised if the engineer inspects 20 buildings is 3.
Step-by-step explanation:
For each building, there are only two possible outcomes after a earthquake. Either they have been damaged, or they have not. The probability of a building being damaged is independent from other buildings. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
15% of all buildings have been structurally compromised.
This means that
20 buildings
This means that
a) What is the probability that if engineers inspect 20 buildings they will find exactly one that is structurally compromised?
This is P(X = 1).
13.68% probability that if engineers inspect 20 buildings they will find exactly one that is structurally compromised.
b) What is the probability that if engineers inspect 20 buildings they will find less than 2 that are structurally compromised?
17.56% probability that if engineers inspect 20 buildings they will find less than 2 that are structurally compromised
c) What is the probability that if engineers inspect 20 buildings they will find greater than 4 that are structurally compromised?
Either they find 4 or less, or they find more than 4. The sum of the probabilities of these events is 1. So
In which
17.02% probability that if engineers inspect 20 buildings they will find greater than 4 that are structurally compromised
d) What is the probability that if engineers inspect 20 buildings they will find between 2 and 5 (inclusive) that are structurally compromised?
75.70% probability that if engineers inspect 20 buildings they will find between 2 and 5 (inclusive) that are structurally compromised
e) What is the expected number of buildings that an engineer will find structurally compromised if the engineer inspects 20 buildings?
The expected value of the binomial distribution is:
So
The expected number of buildings that an engineer will find structurally compromised if the engineer inspects 20 buildings is 3.
Answer:
Step-by-step explanation:
You can find your answer in attached document.
Alexis worked for 59 hours, Becca worked for 74 hours and Cindy worked for 152 hours.
Step-by-step explanation:
Given,
Total hours worked by three of them = 285 hours
Let,
x represent the hours of Alexis.
y represent the hours of Becca.
z represent the hours of Cindy.
According to given statement;
x+y+z=285 Eqn 1
Alexis worked 15 hours less than Becca.
x = y-15 Eqn 2
z = 2y+4 Eqn 3
Putting value of x and z from Eqn 2 and 3 in Eqn 1
Dividing both sides by 4
Putting y=74 in Eqn 2
Putting y=74 in Eqn 3
Alexis worked for 59 hours, Becca worked for 74 hours and Cindy worked for 152 hours.
Keywords: linear equation, division
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