Answer:
usual number of yellow eggs = 12
Usual maximum = 21
Usual minimum = 3
Step-by-step explanation:
To solve this, we will use the expected value of a binomial probability.
The formula is;
E(X) = np
Where;
n is sample size
p is probability of success.
We are given;
n = 58
p = 21%
Thus;
usual number of yellow eggs in samples = np = 58 × 21% = 12.18 ≈ 12
From USL(Upper specification limit) and LSL(Lower specification limit) formula, we can find the maximum usual number and minimum usual number of eggs respectively.
Thus;
USL = n(p + 3√(p(1 - p)/n)
USL = 58(0.21 + 3√(0.21(1 - 0.21)/58)
USL = 21.48 ≈ 21
LSL = n(p - 3√(p(1 - p)/n)
LSL = 58(0.21 - 3√(0.21(1 - 0.21)/58)
LSL = 2.87 ≈ 3
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Answer:
60
Step-by-step explanation:
60* of the circumference would be 60-the total circumference.
(x−a)(x−b)
=(x+−a)(x+−b)
=(x)(x)+(x)(−b)+(−a)(x)+(−a)(−b)
=x2−bx−ax+ab
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=ab−ax−bx+x2</em></u></h2>
