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:
D. 15×4/100×4 = 60/400
Step-by-step explanation:
First I figured out how many photos there are in total, which was by doing 60/.15, since only 15% of the total photos are in black and white. We can already cross out all possibilities that don't have the denominator 400, but we can go a step further by seeing which one is 60/400, or equal to 60/400, which is D. 15×4/100×4 = 60/400.
the value of 25% marked up on $6.75
HI, please send the results from part C….. We can’t answer without knowing the results unless someone already did this question and know the answer..
68.48
100% divided by 6.25% is 16 columns
Second row is 4.28 each
4.28(16)=$68.48