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
Answer:
z = (17+4w)/(w-m)
Step-by-step explanation:
w • (-4+ z) = mz + 17
Distribute
-4w +wz = mz+17
Subtract mz from each side
-4w +wz - mz = mz+17-mz
-4w +wz-mz = 17
Add 4w to each side
-4w +4w+wz-mz = 17+4w
wz-mz = 17+4w
Factor out z
z(w-m) = 17+4w
Divide by (w-m)
z(w-m)/(w-m) = (17+4w)/(w-m)
z = (17+4w)/(w-m)
Answer:
(2n-1)(n-5)
Step-by-step explanation:
2n2-11n+5
2n2-10n-1n+5
(factorize)
2n(n-5)-1(n-5)
(2n-1)(n-5)
Answer:
A
Step-by-step explanation:
m=(y2-y1)/(x2-x1)
Therefore:
(-3-(-4))/(-5-1)
-1/6 (A)
