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
Area of the box:
A = 2 a² + 4 a h = 600 cm²
h = a ( the box is a cube )
A = 6 a² = 600 cm²
a² = 100
a = 10 cm
V = a³ = 10³ = 1000 cm³
To solve this problem you must apply the proccedure shown below:
1. You have that the following information given in the problem above: The<span> photograph is reduced by a scale factor of 3/8 and the original had a length of 20 inches,
2- Therefore you have:
length=20 inches-</span>(20 inches x 3/8)<span>
length=20 inches-7.5 inches
length=12.5 inches
3- Then, as you can see, the answer is: 12.5 inches.</span>
Slopt in form ax+by=c is -a/b
-3/3=-1
slope is -1
intercept in form ax+by=c is c/b
63/3=9
slope is -1
intercept is at y=9