Answer:

Step-by-step explanation:
(32 x 32) = 32^2
<u>solution</u>
W=f×d×cos theta
=25N×16m×cos 45°
=400Nm×cos 45°
=282.84J
W=f×d×cos theta
=25N×7m×cos 10°
=175Nm×cos10°
=172.34J
Total workdone=282.84J+172.34J=455.18J
The correct answer is A: 3. This is because...
3 * 4 = 12
12 * 5 = 60
Therefore, the answer is 60.
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