Answer:
Step-by-step explanation:
2 l + 2w = 34 ⇔ l + w = 17 ⇒ l = 17 - w ..... <em>(1)</em>
l w = 66 ..... <em>(2)</em>
<em>(1)</em> ----> <em>(2)</em>
(17 - w) w = 66
w² - 17w + 66 = 0 ⇔ (w - 6)(w - 11) = 0
w = 6 cm
l = 11 cm
I hope this helps you
14= 2/3 (9y-15) divided 2 each sides
7= 9y-15/3
7.3= 9y-15
21+15= 9y
36= 9y
y= 4
Answer:
Probability that a randomly selected broiler weighs more than 1454 g is 0.3372 or 34% (approx.)
Step-by-step explanation:
Given:
Weights of Broilers are normally distributed.
Mean = 1387 g
Standard Deviation = 161 g
To find: Probability that a randomly selected broiler weighs more than 1454 g.
we have ,


X = 1454
We use z-score to find this probability.
we know that


P( z = 0.42 ) = 0.6628 (from z-score table)
Thus, P( X ≥ 1454 ) = P( z ≥ 0.42 ) = 1 - 0.6628 = 0.3372
Therefore, Probability that a randomly selected broiler weighs more than 1454 g is 0.3372 or 34% (approx.)