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
A=5.00
B=7.07
I hope this helps you out
C. 2.4 x 10^6
D. 1.47 x 10^-3
I hope this helps !
Infinite solution Bc both sides are equal
Answer:
ok
Step-by-step explanation:
