Answer:
The probability that the weight of a randomly selected steer is between 920 and 1730 lbs
<em> P(920≤ x≤1730) = 0.7078 </em>
Step-by-step explanation:
<u><em>Step(i):-</em></u>
Given mean of the Population = 1100 lbs
Standard deviation of the Population = 300 lbs
Let 'X' be the random variable in Normal distribution
Let x₁ = 920

Let x₂ = 1730

<u><em>Step(ii)</em></u>
The probability that the weight of a randomly selected steer is between 920 and 1730 lbs
P(x₁≤ x≤x₂) = P(Z₁≤ Z≤ Z₂)
= P(-0.6 ≤Z≤2.1)
= P(Z≤2.1) - P(Z≤-0.6)
= 0.5 + A(2.1) - (0.5 - A(-0.6)
= A(2.1) +A(0.6) (∵A(-0.6) = A(0.6)
= 0.4821 + 0.2257
= 0.7078
<u><em>Conclusion:-</em></u>
The probability that the weight of a randomly selected steer is between 920 and 1730 lbs
<em> P(920≤ x≤1730) = 0.7078 </em>
Answer:
the green kin quadrant IV
Step-by-step explanation:
i took the quiz and got it right
Answer:
x-int is 3
y-int is -4
Step-by-step explanation:
when x is zero, -9y = 36
y = -4
when y = zero 12x = 36
x = 3