Answer:
The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors
P(X⁻ ≥ 96.3) = 0.0087
Step-by-step explanation:
<u><em>Step(i):-</em></u>
Given that the mean of the Population = 95
Given that the standard deviation of the Population = 5
Let 'X' be the random variable in a normal distribution
Let X⁻ = 96.3
Given that the size 'n' = 84 monitors
<u><em>Step(ii):-</em></u>
<u><em>The Empirical rule</em></u>
Z = 2.383
The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors
P(X⁻ ≥ 96.3) = P(Z≥2.383)
= 1- P( Z<2.383)
= 1-( 0.5 -+A(2.38))
= 0.5 - A(2.38)
= 0.5 -0.4913
= 0.0087
<u><em>Final answer:-</em></u>
The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors
P(X⁻ ≥ 96.3) = 0.0087
Bag of potatoes are usually 10 pounds
I would say loneliness, because she is taken by a kings men .. and it’s not him
Explanation:
<u>Urgent care centers</u>: care for basic needs after regular doctor hours
<u>Hospitals</u>: treat time-sensitive emergencies
<u>Medical specialists</u>: offer treatment in a specific field of medicine, such as cardiology
<u>General practice doctors and nurse practitioners</u>: care for routine medical needs
<u>Crisis pregnancy centers</u>: provide counseling for unplanned pregnancies
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<em>Discussion</em>
Urgent care centers are often open all hours, but may not be as fully equipped (or staffed) to provide the sort of emergency medicine that a fully-equipped hospital can provide. While a general- or nurse-practitioner can provide routine care, they will consult with specialists when expertise is needed in a specific area.
Various kinds of pregnancy centers can provide counseling and perhaps some medical services for planned or unplanned pregnancies.
Answer:
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Step-by-step explanation:
