Answer:
the probability of no defects in 10 feet of steel = 0.1353
Step-by-step explanation:
GIven that:
A roll of steel is manufactured on a processing line. The anticipated number of defects in a 10-foot segment of this roll is two.
Let consider β to be the average value for defecting
So;
β = 2
Assuming Y to be the random variable which signifies the anticipated number of defects in a 10-foot segment of this roll.
Thus, y follows a poisson distribution as number of defect is infinite with the average value of β = 2
i.e
the probability mass function can be represented as follows:
where;
y = 0,1,2,3 ...
Hence, the probability of no defects in 10 feet of steel
y = 0
P(y =0) = 0.1353
A = L * W
A = 320
L = W + 4
320 = W(W + 4)
320 = W^2 + 4W
W^2 + 4W = 320
W^2 + 4W + 4 = 320 + 4
(W + 2)^2 = 324
W + 2 = (+-) sqrt 324
W = -2 (+-) 18
W = -2 + 18 = 16 ft <== this is the width
W = -2 - 18 = -20....not this one because it is negative
L = W + 4
L = 16 + 4
L = 20 ft <=== this is the length
in summary...the width is 16 ft and the length is 20 ft
Answer:
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Step-by-step explanation:
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<u>Answer:</u>
- The simplified expression is "7p/4 + 2 1/2" or "1.75p + 2.5".
<u>Step-by-step explanation:</u>
- 2p + 3/4p + 6 - p - 3 1/2
- => (2p - p + 3/4p) + (6 - 3.5)
- => 1.75p + 2.5
Hence, the simplified expression is "<u>7p/4 + 2 1/2</u>" or "<u>1.75p + 2.5</u>". Any of these would work.
Hoped this helped.
Answer:
a.
<u>Required equation is</u>
b.
<u>Solve the equation</u>
