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Nuetrik [128]
3 years ago
14

Complete the patterns 0.02 0.2 blank 20 blank blank

Mathematics
2 answers:
katen-ka-za [31]3 years ago
8 0

The Complete pattern is; 0.02, 0.2, 2, 20, 200, 2000

<h2>Further Explanation;</h2><h3>Sequence</h3>
  • A sequence refers to a set of numbers, called terms, arranged in some particular order.
  • There are two major types of sequences.

Arithmetic sequence

  • An arithmetic sequence is a sequence with the difference between two consecutive terms constant. The difference is called the common difference (d).

For example;

2,4,6,8,10,12,.............. is an arithmetic sequence with a common difference of 2.

  • The nth term of an arithmetic series is therefore given by;

a_{n} =a_{1} + (n-1)d

Where a_{1} is the first term and d is the common difference.

<h3>Geometric sequence</h3>
  • A geometric sequence is a sequence with the ratio between two consecutive terms constant.
  • The sequence follows a pattern where the next term is found by multiplying by a constant called the common ratio, r.

For example;

0.02, 0.2, 2, 20, 200, 2000, ........... is a geometric sequence in which the common ratio is 10.

The nth term of a geometric sequence is given by;

a_{n} =a_{1} r^{n-1}

where a_{1} is the first term and r is the common ratio

Key words: sequence, arithmetic sequence, geometric sequence, common difference, common ratio

<h3>Learn more about;</h3>
  • Arithmetic sequence: brainly.com/question/11718753
  • Geometric sequence: brainly.com/question/9300199
  • Example of a geometric sequence: brainly.com/question/10601401

Level: High school

Subject: Mathematics

Topic: Sequence and series

Sub-topic: Geometric sequence

andrew-mc [135]3 years ago
5 0
Just multiply the number by 10
0.02
0.2
2
20
200
2000
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Select all the factors roots of x^3 +2x^2-16x-32
Ivanshal [37]

Factor by grouping. Group up the terms into pairs, factor each pair, then factor out the overall GCF.

x^3 + 2x^2 - 16x - 32

(x^3 + 2x^2) + (-16x-32) ... pair up terms

x^2(x + 2) + (-16x - 32) ... factor x^2 from the first group

x^2(x + 2) - 16(x + 2) ... factor -16 from the second group

(x^2 - 16)(x + 2) .... factor out (x+2)

(x - 4)(x + 4)(x + 2) .... Use the difference of squares to factor x^2-16

---------------------------

The original expression completely factors to (x - 4)(x + 4)(x + 2)

The three factors are x - 4  and  x + 4   and   x + 2

5 0
3 years ago
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Is 11/20 greater than a half
Rainbow [258]
Yes... a half is equal to 10/20 and 11/20 is greater than 10/20 so 11/20 is greater than a half
8 0
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Let $$X_1, X_2, ...X_n$$ be uniformly distributed on the interval 0 to a. Recall that the maximum likelihood estimator of a is $
Solnce55 [7]

Answer:

a) \hat a = max(X_i)  

For this case the value for \hat a is always smaller than the value of a, assuming X_i \sim Unif[0,a] So then for this case it cannot be unbiased because an unbiased estimator satisfy this property:

E(a) - a= 0 and that's not our case.

b) E(\hat a) - a= \frac{na}{n+1} - a = \frac{na -an -a}{n+1}= \frac{-a}{n+1}

Since is a negative value we can conclude that underestimate the real value a.

\lim_{ n \to\infty} -\frac{1}{n+1}= 0

c) P(Y \leq y) = P(max(X_i) \leq y) = P(X_1 \leq y, X_2 \leq y, ..., X_n\leq y)

And assuming independence we have this:

P(Y \leq y) = P(X_1 \leq y) P(X_2 \leq y) .... P(X_n \leq y) = [P(X_1 \leq y)]^n = (\frac{y}{a})^n

f_Y (Y) = n (\frac{y}{a})^{n-1} * \frac{1}{a}= \frac{n}{a^n} y^{n-1} , y \in [0,a]

e) On this case we see that the estimator \hat a_1 is better than \hat a_2 and the reason why is because:

V(\hat a_1) > V(\hat a_2)

\frac{a^2}{3n}> \frac{a^2}{n(n+2)}

n(n+2) = n^2 + 2n > n +2n = 3n and that's satisfied for n>1.

Step-by-step explanation:

Part a

For this case we are assuming X_1, X_2 , ..., X_n \sim U(0,a)

And we are are ssuming the following estimator:

\hat a = max(X_i)  

For this case the value for \hat a is always smaller than the value of a, assuming X_i \sim Unif[0,a] So then for this case it cannot be unbiased because an unbiased estimator satisfy this property:

E(a) - a= 0 and that's not our case.

Part b

For this case we assume that the estimator is given by:

E(\hat a) = \frac{na}{n+1}

And using the definition of bias we have this:

E(\hat a) - a= \frac{na}{n+1} - a = \frac{na -an -a}{n+1}= \frac{-a}{n+1}

Since is a negative value we can conclude that underestimate the real value a.

And when we take the limit when n tend to infinity we got that the bias tend to 0.

\lim_{ n \to\infty} -\frac{1}{n+1}= 0

Part c

For this case we the followng random variable Y = max (X_i) and we can find the cumulative distribution function like this:

P(Y \leq y) = P(max(X_i) \leq y) = P(X_1 \leq y, X_2 \leq y, ..., X_n\leq y)

And assuming independence we have this:

P(Y \leq y) = P(X_1 \leq y) P(X_2 \leq y) .... P(X_n \leq y) = [P(X_1 \leq y)]^n = (\frac{y}{a})^n

Since all the random variables have the same distribution.  

Now we can find the density function derivating the distribution function like this:

f_Y (Y) = n (\frac{y}{a})^{n-1} * \frac{1}{a}= \frac{n}{a^n} y^{n-1} , y \in [0,a]

Now we can find the expected value for the random variable Y and we got this:

E(Y) = \int_{0}^a \frac{n}{a^n} y^n dy = \frac{n}{a^n} \frac{a^{n+1}}{n+1}= \frac{an}{n+1}

And the bias is given by:

E(Y)-a=\frac{an}{n+1} -a=\frac{an-an-a}{n+1}= -\frac{a}{n+1}

And again since the bias is not 0 we have a biased estimator.

Part e

For this case we have two estimators with the following variances:

V(\hat a_1) = \frac{a^2}{3n}

V(\hat a_2) = \frac{a^2}{n(n+2)}

On this case we see that the estimator \hat a_1 is better than \hat a_2 and the reason why is because:

V(\hat a_1) > V(\hat a_2)

\frac{a^2}{3n}> \frac{a^2}{n(n+2)}

n(n+2) = n^2 + 2n > n +2n = 3n and that's satisfied for n>1.

8 0
4 years ago
A drawer has a volume of 1575 cubic inches. the drawer is 5 inches deep and 21 inches long. What is the width of the drawer?
katrin [286]
Volume is l × w × h
l = 21 in
h = 5 in
w = ?
volume = 1575 in

21 × 5 × w = 1575
105 × w = 1575
w = 1575 ÷ 105
w = 15 inches
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3 years ago
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nignag [31]
Simplify <span>-6(1+11b) using distributive property

-6 - 66b</span>
6 0
3 years ago
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