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Dmitrij [34]
2 years ago
8

List the first 10 terms of each of these sequences.

Mathematics
2 answers:
harina [27]2 years ago
7 0
Q1)
the sequence should start with 10, after that each term is calculated by subtracting 3 from the previous term.
1st term - 10
2nd term - 10 - 3 = 7
3rd term  - 7 - 3 = 4
4th term - 4 - 3 = 1
5th term  - 1 - 3 = -2
6th term - -2 - 3 = -5
7th term - -5 - 3 = -8 
8th - -8 - 3 = -11
9th - -11 - 3  = -14
10th -14 - 3 = -17
the sequence is - 10,7,4,1,-2,-5,-8,-11,-14,-17

Q2) 
<span>the sequence whose nth term is the sum of the first n positive integers
In this we get the term by adding all the integers of the terms before that term 
1st term - n = 1 no terms before this , therefore 0 + n(1)  = 1
2nd term -n =2  sum of integers before - 1 + n( 2) = 3
3rd - 3+3 = 6
4th - 6+4 = 10 
5th - 10 + 5 = 15
6th - 15 + 6 = 21
7th  - 21 + 7 = 28
8th - 28 + 8 = 36
9th - 36 + 9 = 45
10th -  45 + 10 = 55
this is a triangular number pattern
this number pattern can be found out using ; n = (n x (n+1))/2

sequence is - 1,3,6,10,15,21,28,36,45,55

Q3)
</span>the sequence whose nth term is 3n − 2n
general term for this sequence is 3n − 2n
to find 1st term , n = 1
substituting n = 1 in the general term 
1st term - 3x1 - 2x1 = 3-2 = 1
2nd - 3x2- 2x2 = 6 - 4 = 2
3rd - 3x3 - 2x3 = 9-6 = 3
4th - 3x4 - 2x4 = 12 - 8 = 4
5th - 3x5 - 2x5 = 15 - 10 = 5
6th - 3x6 - 2x6 = 18 - 12 = 6
7th - 3x7 - 2x7 = 21 - 14 = 7
8th - 3 x8 - 2x8 = 24 - 16 = 8
9th - 3x9 - 2x9 = 27 - 18 = 9
10th - 3x10 - 2x10 = 30-20 = 10

sequence is 1,2,3,4,5,6,7,8,9,10

Q4)
<span>the sequence whose nth term is √ n
when n=1 1st term is </span>√1 = 1
1st term - √1 = 1
2nd term - √2 = 1.41
3rd - √3 = 1.73
4th - √4 = 2
5th - √5 = 2.23
6th- √6 = 2.44
7th - √7 = 2.65
8th- √8 = 2.82
9th - √9 = 3
10th - √10 = 3.16

The sequence is 1, 1.41, 1.73, 2, 2.23, 2.44, 2.65, 2.82, 3, 3.16

Q5)T<span>he sequence whose first two terms are 1 and 5 and each succeeding term is the sum of the two previous terms
</span>1st term - 1
2nd term - 5
3rd term - add 1st and 2nd term (1+5) = 6
4th term - add 2nd and 3rd terms (5+6) = 11
5th - add 3rd and 4th (6+11) = 17
6th - (11+17) = 28
7th - (17 + 28) = 45
8th - 45 + 28 = 73
9th - 73 + 45 = 118
10th - 73+ 118 = 191

sequence is - 1,5,6,11,17,28,45,73,118,191
Semenov [28]2 years ago
7 0
A. We start by writting down the general formula:
a(1) = 10
a(2) = a(1) - 3  
a(3) = a(2) - 3
a(n) = a(n-1) -3

Now we find first ten terms:
10, 7, 4 ,1, -2, -5, -8, -11, -14, -17

b. We start by writting down the general formula:
a(1) = 1
a(2) = 1 + 2
a(3) = 1 + 2 + 3
a(n) = 1 + 2 + ... + n

Now we find first ten terms:
1, 3, 6, 10, 15, 21, 28, 36, 45, 55

c. We start by writting down the general formula:
a(n) = 3n -2n = n

Now we find first ten terms:
1, 2, 3, 4, 5, 6, 7, 8, 9, 10

d. We start by writting down the general formula:
a(n)= \sqrt{n}

Now we find first ten terms:
1,  \sqrt{2},\sqrt{3}, 2,\sqrt{5},\sqrt{6},\sqrt{7},\sqrt{8},3,\sqrt{10}

e. We start by writting down the general formula:
a(1) = 1
a(2) = 5
a(3) = a(1) + a(2)
a(4) = a(2) + a(3)
a(n) = a(n-2) + a(n-1)

Now we find first ten terms:
1, 5, 6, 11, 17, 28, 45, 73, 118, 191
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Finger [1]

Answer:

79.5 + 5.5x = Y

Step-by-step explanation:

Sumo wrestler gained 5.5 kg per month

After 11 month, he weighed 140 kg.

Let x be his current weight.

Then x + 11(5.5) = 140

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If Y is the weight of the wrestler after t months, then the linear equation would be:

 79.5 + 5.5t = Y

5 0
3 years ago
(A) A small business ships homemade candies to anywhere in the world. Suppose a random sample of 16 orders is selected and each
Leviafan [203]

Following are the solution parts for the given question:

For question A:

\to (n) = 16

\to (\bar{X}) = 410

\to (\sigma) = 40

In the given question, we calculate 90\% of the confidence interval for the mean weight of shipped homemade candies that can be calculated as follows:

\to \bar{X} \pm t_{\frac{\alpha}{2}} \times \frac{S}{\sqrt{n}}

\to C.I= 0.90\\\\\to (\alpha) = 1 - 0.90 = 0.10\\\\ \to \frac{\alpha}{2} = \frac{0.10}{2} = 0.05\\\\ \to (df) = n-1 = 16-1 = 15\\\\

Using the t table we calculate t_{ \frac{\alpha}{2}} = 1.753  When 90\% of the confidence interval:

\to 410 \pm 1.753 \times \frac{40}{\sqrt{16}}\\\\ \to 410 \pm 17.53\\\\ \to392.47 < \mu < 427.53

So 90\% confidence interval for the mean weight of shipped homemade candies is between 392.47\ \ and\ \ 427.53.

For question B:

\to (n) = 500

\to (X) = 155

\to (p') = \frac{X}{n} = \frac{155}{500} = 0.31

Here we need to calculate 90\% confidence interval for the true proportion of all college students who own a car which can be calculated as

\to p' \pm Z_{\frac{\alpha}{2}} \times \sqrt{\frac{p'(1-p')}{n}}

\to C.I= 0.90

\to (\alpha) = 0.10

\to \frac{\alpha}{2} = 0.05

Using the Z-table we found Z_{\frac{\alpha}{2}} = 1.645

therefore 90\% the confidence interval for the genuine proportion of college students who possess a car is

\to 0.31 \pm 1.645\times \sqrt{\frac{0.31\times (1-0.31)}{500}}\\\\ \to 0.31 \pm 0.034\\\\ \to 0.276 < p < 0.344

So 90\% the confidence interval for the genuine proportion of college students who possess a car is between 0.28 \ and\ 0.34.

For question C:

  • In question A, We are  90\% certain that the weight of supplied homemade candies is between 392.47 grams and 427.53 grams.
  • In question B, We are  90\% positive that the true percentage of college students who possess a car is between 0.28 and 0.34.

Learn more about confidence intervals:  

brainly.in/question/16329412

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2 years ago
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