Answer:
p
Step-by-step explanation:
Using the normal distribution, it is found that there is a 0.2776 = 27.76% probability that the life span of the monitor will be more than 20,179 hours.
<h3>Normal Probability Distribution</h3>
The z-score of a measure X of a normally distributed variable with mean
and standard deviation
is given by:

- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
The mean and the standard deviation are given, respectively, by:

The probability that the life span of the monitor will be more than 20,179 hours is <u>one subtracted by the p-value of Z when X = 20179</u>, hence:


Z = 0.59.
Z = 0.59 has a p-value of 0.7224.
1 - 0.7224 = 0.2776.
0.2776 = 27.76% probability that the life span of the monitor will be more than 20,179 hours.
More can be learned about the normal distribution at brainly.com/question/24663213
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Answer:
Idk if this is right but I think it's The transformed shaped shifted 7 units to the right so add 7 to x and it shifted 6 units down so subtract 6 from y.
The answer is (x,y) = (x+7) (y-6)
Hope this helps!!!?
Find 10% of it (just move the decimal point one spot to the left) and multiply by 4
Answer: a - 4.512 hours
b - 1.94 hours
Step-by-step explanation:
Given,
a) A(t) = 10 (0.7)^t
To determine when 2mg is left in the body
We would have,
A(t) = 2, therefore
2 = 10(0.7)^t
0.7^t =2÷10
0.7^t = 0.2
Take the log of both sides,
Log (0.7)^t = log 0.2
t log 0.7 = log 0.2
t = log 0.2/ 0.7
t = 4.512 hours
Thus it will take 4.512 hours for 2mg to be left in the body.
b) Half life
Let A(t) = 1/2 A(0)
Thus,
1/2 A(0) = A(0)0.7^t
Divide both sides by A(0)
1/2 = 0.7^t
0.7^t = 0.5
Take log of both sides
Log 0.7^t = log 0.5
t log 0.7 = log 0.5
t = log 0.5/log 0.7
t = 1.94 hours
Therefore, the half life of the drug is 1.94 hours