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
Answer:
e : g = 2 : 7
Step-by-step explanation:
e/f = 3/7
e = 3f/7
f/g = 2/3
g = 3f/2
e/g = (3f/7)/(3f/2) = 2/7
I've attached the graphs to this answer. I hope they help.
Answer:
false
Step-by-step explanation:
any line segment or distance has only one midpoint.
there is only one middle to everything.
how would that look like, if something had 2 middles ?
then they would not be a middle, and the middle between these 2 middles would then be the real middle ...
Option 2. Is the answer to ur question..!!
Hope this helps you!!