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: <em>81</em>
Step-by-step explanation:
<em>3^4</em>
<em>3x3x3x3</em>
<em>3x3=9</em>
<em>9x3=27</em>
<em>27x3=</em><em>81</em>
Answer:
Step-by-step explanation:
The answer would be 15 +20x. You have to combine like terms.
2+6+7+20x
8+7+20x
= 15+ 20x (answer)
B because it's not putting the blame on anyone. It is completely unbiased and objective.