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:
48 miles per hour
Step-by-step explanation:
If they traveled 2400 miles and the total driving time was 50 hours, divide 2400/50 in order to find how many miles they drove per hour.
John will pay $8.68 for the combined cost of 1 pound of banana and 1 pound of cherries.
Let: b=cost of banana per pound and c=cost of cherries per pound
Equation 1: For 3 pounds of cherries and 2 pounds of bananas, John pays a total of $24.95.
3c + 2b =$24.95
Equation 2: The cost of bananas is $6.50 less than a pound of cherries.
b= c - $6.50
We can substitute the second equation into the first one to solve for the cost of cherries per pound.
3c + (2)(c-$6.50)= $24.95
3c + 2c -$13.00 = $24.95
5c = $24.95 + $13.00
c = $7.59
Substituting the value of c to the second equation to solve for b.
b= $7.59 - $6.50 = $1.09
The combined cost of 1 pound of banana and 1 pound of cherries is $1.09 + $7.59 or $8.68.
For more information regarding the system of equations, please refer to brainly.com/question/25976025.
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Answer:
3x + x + 3x + x = 360
8x = 360
x = 45
3(45)= 135
Step-by-step explanation:
Answer:
.14
Step-by-step explanation: