Because P(A n B) has been defined as x, and we calculated the value of P(A n B) = P(A) P(B) to be x² + 0.49x + 0.058, we now solve for x, the probability of both pumps failing. P(An B) = P(A) P(B) x = x² + 0.49x + 0.058 0 = x² - (C × x + 0.058 Solving the quadratic equation for x results in two values that represent the possible probabilities that the pumping. system will fail on any given day. Rounding both values to four decimal places, the lower value is x = the larger is x = and

Intermediate Algebra
10th Edition
ISBN:9781285195728
Author:Jerome E. Kaufmann, Karen L. Schwitters
Publisher:Jerome E. Kaufmann, Karen L. Schwitters
Chapter11: Exponential And Logarithmic Functions
Section11.CR: Review Problem Set
Problem 62CR
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Because P(A n B) has been defined as x, and we calculated the value of P(An B) = P(A) · P(B) to be
x² + 0.49x + 0.058, we now solve for x, the probability of both pumps failing.
P(A n B) = P(A) · P(B)
x = x² + 0.49x + 0.058
0 = : x² - (
Solving the quadratic equation for x results in two values that represent the possible probabilities that the pumping
system will fail on any given day. Rounding both values to four decimal places, the lower value is x =
the larger is x =
|× )×·
+ 0.058
, and
Transcribed Image Text:Because P(A n B) has been defined as x, and we calculated the value of P(An B) = P(A) · P(B) to be x² + 0.49x + 0.058, we now solve for x, the probability of both pumps failing. P(A n B) = P(A) · P(B) x = x² + 0.49x + 0.058 0 = : x² - ( Solving the quadratic equation for x results in two values that represent the possible probabilities that the pumping system will fail on any given day. Rounding both values to four decimal places, the lower value is x = the larger is x = |× )×· + 0.058 , and
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