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Mademuasel [1]
2 years ago
12

The sum of the first 15 terms of the sequence an=10n+21

Mathematics
1 answer:
enot [183]2 years ago
3 0
That is
the sum of all the terms from a1 to an when you are summing up n terms is n(a1+an)/2
so

find the 1st and 15th term
1st term is a1=10(1)=21=10+21=31
15th term is a15=10(15)+21=150+21=171

and n=15
15(31+171)/2=15(202)/2=3030/2=1515
the sum of the first 15 term is 1515
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Standard deviation = \sigma=\sqrt{np(1-p)}

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Using the standard normal distribution table ,

P(x=4)=P(x\leq4)-P(x\leq3)              (1)

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z=\dfrac{4-1.8}{1.2369}\approx1.78

For x=3

z=\dfrac{3-1.8}{1.2369}\approx0.97

Then , from (1)

P(x=4)=P(z\leq1.78)-P(z\leq0.97)\\\\=0.962462-0.8339768\approx0.1289    

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In 154 N38 what digit should be replaced to N to make the number divisible by 3,6 and 9?​
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Answer:

N = 6 so that number becomes divisible by 3, 6 and 9.

Step-by-step explanation:

In Number Theory there is a rule of thumb which states that sum of digits of a multiple of 3 equal 3 or a multiple of three. If we know that n = 154N38, then its sum of digits is:

x = 1 + 5+4+N+3+8

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We get the following four distinct options: 154038, 154338, 154638, 154938. Now we find which number is divisible by 6 and 9 by factor decomposition:

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It is quite evident that N = 6 so that number becomes divisible by 3, 6 and 9.

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