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Maksim231197 [3]
3 years ago
9

Systematic Sampling Technique is used to select 5 numbers between 1 and 200. The first selected number is 12. What should be nex

t four numbers?
Mathematics
1 answer:
Zina [86]3 years ago
3 0

Answer:  52, 92 , 132, 172

Step-by-step explanation:

  • A systematic random sampling is a random sampling technique in which a sample from a large population are chosen according to a random beginning point and a fixed periodic interval (k).

Given : Systematic Sampling Technique is used to select 5 numbers between 1 and 200.

⇒ Population size : N=200

Required Sample size  : n=5

Since , Sampling interval is given by :-

k=\dfrac{N}{n}

⇒ [tex]k=\dfrac{200}{5}[=40/tex]

If the first selected number is 12, then the next four numbers would be :

(12+K) , (12+2k) , (12+3k) , (12+4k)

Put value of k , we get

(12+40) , (12+2(40)) , (12+3(40)) , (12+4(40)) =(52, 92 , 132, 172)

Hence, the  next four numbers are : 52, 92 , 132, 172

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GenaCL600 [577]
<h3>Let the number be x</h3>

x:11 =  \frac{1}{3} :6

  • Divide the numbers

\frac{x}{11}  =  \frac{ \frac{1}{3} }{6}  \\  \frac{x}{11}  =  \frac{1}{18}

  • Multiply all terms by the same value to eliminate fraction denominators

11 \times  \frac{x}{11}  = 11 \times  \frac{1}{18}

  • Cancel multiplied terms that are in the denominator

x = 11 \times  \frac{1}{18}

  • Multiply the numbers

x =  \frac{11}{18}

<h2>Solved ✔︎</h2>

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3 years ago
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Round to nearest centimeter 80.5 cm is how many cm?
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7 0
3 years ago
James folds a piece of paper in half several times,each time unfolding the paper to count how many equal parts he sees. After fo
Snezhnost [94]

Answer:

There will be total 2048 parts of the given paper if James if able to fold the paper eleven times.

The needed function is y = 2 ^n

Step-by-step explanation:

Let us assume the piece of paper James decides to fold is a SQUARE.

Now, let us assume:

n : the number of times the paper is folded.

y : The number of parts obtained after folds.

Now, if the paper if folded ONCE ⇒  n = 1

Also, when the pap er is folded once, the parts obtained are TWO equal parts.

⇒  for n = 1 , y = 2       ..... (1)

Similarly, if the paper if folded TWICE  ⇒  n = 2

Also, when the paper is folded twice, the parts obtained are FOUR equal parts.

⇒  for n = 2 , y = 4       ..... (2)

⇒y  = 2^2  =  2^n

Continuing the same way, if the paper is folded SEVEN times  ⇒  n = 7

So, y = 2^ n = 2^7 = 128

⇒  There are total 128 equal parts.

Lastly,  if the paper is folded ELEVEN  times  ⇒  n = 11

So, y = 2^ n = 2^{11} = 2048

⇒  There are total 2048 equal parts.

Hence, there will be total 2048 parts of the given paper if James if able to fold the paper eleven times.

And the needed function is y = 2 ^n

8 0
4 years ago
Please some one help me.​
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Answer:

-11

Step-by-step explanation:

You need to take the absolute value of 12-15 which would be 3. Then take -8-3 which would be -11

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Answer:

0

Step-by-step explanation:

The answer would be 0

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