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:
-45
Step-by-step explanation:
Two intersecting line segments forming an angle marked with a small box.
The "small box" represents 90°, and perpendicular lines are 2 intersecting lines that form 90° angles
Given:
n = 50, sample size
, sample mean
s = 2.4 min, sample standard deviation.
The confidence interval is
At the 99% confidence level, t* (from the student's t-distribution) is
t* = 2.68
Therefore
t*(s/√n) = 2.68*(2.4/√50) = 0.9096
The confidence interval is
(23.6-0.9096, 23.6+0.9096) ≈ (22.69, 24.51)
Answer: (22.7, 24.5)
The domain for N is All integers where n ≥ 1
<u>Solution:</u>
According to statement a1 = 2 and r = 4. This shows that r is greater than 1.
If r is greater than 1 than it includes integers greater than 1 or equal to 1. It does not include all the real numbers because real numbers include negative numbers also.
If starting value is 2, if we put n=0, then we get 2, but if we put a negative value than we would get a number which is not a part of our sequence.
Thus the domain of n is All integers where n greater than or equal to 1
