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zhuklara [117]
3 years ago
9

Write the first five terms of the sequence in which the nth term is a_n=(n+1)!/n+1.

Mathematics
2 answers:
Aleksandr [31]3 years ago
6 0

Answer: a) 1,2,6,24,120

Step-by-step explanation:

VikaD [51]3 years ago
4 0

Answer:a) 1,2,6,24,120

Step-by-step explanation:

The formula for the nth term of the sequence is expressed as

n=(n+1)!/n+1

For the first term, n = 1. Therefore,

(1+1)!/(1+1) = 2!/2 = (2×1)/2 = 1

For the second tern, n = 2. Therefore,

(2+1)!/2+1 = 3! /3 = (3×2×1)/3 = 6/3 = 2

For the third term, n = 3, therefore

(3+1)!/3+1 = 4!/4 =(4×3×2×1)/4 = 6

For the fourth term, n = 4. Therefore

(4+1)!/4+1 = 5!/5 = (5×4×3×2×1)/5 = 24

For the fifth term, n = 5. Therefore

(5+1)!/5+1 = 6!/6 = 6× 5×4×3×2×1)/6 = 120

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5 0
3 years ago
[Q71 Suppose that the height, in inches, of a 25-year-old man is a normal random variable with parameters g = 71 inch and 02 = 6
viktelen [127]

Answer: (a) Percentage of 25 year old men that are above 6 feet 2 inches is 11.5%.

              (b) Percentage of 25 year old men in the 6 footer club that are above 6 feet 5 inches are 2.4%.

Step-by-step explanation:

Given that,

                  Height (in inches) of a 25 year old man is a normal random variable with mean g=71 and variance o^{2} =6.25.

To find:  (a) What percentage of 25 year old men are 6 feet, 2 inches tall

               (b) What percentage of 25 year old men in the 6 footer club are over 6 feet. 5 inches.

Now,

(a) To calculate the percentage of men, we have to calculate the probability

P[Height of a 25 year old man is over 6 feet 2 inches]= P[X>74in]

                           P[X>74] = P[\frac{X-g}{o} > \frac{74-71}{2.5}]

                                         = P[Z > 1.2]

                                         = 1 - P[Z ≤ 1.2]

                                         = 1 - Ф (1.2)

                                         = 1 - 0.8849

                                         = 0.1151

Thus, percentage of 25 year old men that are above 6 feet 2 inches is 11.5%.

(b) P[Height of 25 year old man is above 6 feet 5 inches gives that he is above 6 feet] = P[X, 6ft 5in - X, 6ft]

     P[X > 6ft 5in I X > 6ft] = P[X > 77 I X > 72]

                                          = \frac{P[X > 77]}{P[ X > 72]}

                                          = \frac{P[\frac{X - g}{o}>\frac{77-71}{2.5}]  }{P[\frac{X-g}{o} >\frac{72-71}{2.5}] }

                                          = \frac{P[Z >2.4]}{P[Z>0.4]}

                                          =  \frac{1-P[Z\leq2.4] }{1-P[Z\leq0.4] }

                                          = \frac{1-0.9918}{1-0.6554}

                                          = \frac{0.0082}{0.3446}

                                          = 0.024

Thus, Percentage of 25 year old men in the 6 footer club that are above 6 feet 5 inches are 2.4%.

4 0
3 years ago
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