For (2), start with the base case. When n = 2, we have
(n + 1)! = (2 + 1)! = 3! = 6
2ⁿ = 2² = 4
6 > 4, so the case of n = 2 is true.
Now assume the inequality holds for n = k, so that
(k + 1)! > 2ᵏ
Under this hypothesis, we want to show the inequality holds for n = k + 1. By definition of factorial, we have
((k + 1) + 1)! = (k + 2)! = (k + 2) (k + 1)!
Then by our hypothesis,
(k + 2) (k + 1)! > (k + 2) 2ᵏ = k•2ᵏ + 2ᵏ⁺¹
and k•2ᵏ ≥ 2•2² = 8, so
k•2ᵏ + 2ᵏ⁺¹ ≥ 8 + 2ᵏ⁺¹ > 2ᵏ⁺¹
which proves the claim.
Unfortunately, I can't help you with (3). Sorry!
Answer:
(0.0706, 0.1294)
Step-by-step explanation:
Confidence interval of a proportion is:
CI = p ± CV × SE
where p is the proportion,
CV is the critical value (z score or t score),
and SE is the standard error.
The sample is large enough to estimate as normal. For 95% confidence level, CV = z = 1.96.
Standard error for a proportion is:
SE = √(pq/n)
SE = √(0.1 × 0.9 / 400)
SE = 0.015
The confidence interval is:
CI = 0.1 ± (1.96)(0.015)
CI = (0.0706, 0.1294)
Round as needed.
2/21 is the answer
Hope this helps
7433700 I think but not sure
Yes--- if it is a square BASED pyramid
