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kykrilka [37]
4 years ago
5

Which group of numbers is correctly ordered from least to greatest? -4, -7, -10, 0, 1, 5 0, -4, -1, 6, 11, 14 -10, -7, -4, 0, 1,

9 0, -3, -7, 1, 4, 7
Mathematics
1 answer:
Eduardwww [97]4 years ago
7 0

Numbers are ordered from least to greatest when each number is greater than the number preceding it, and lesser than the number succeeding it.

Rephrasing the answer choices to make it more clear:

-4, -7, -10, 0, 1, 5

-7 is lesser than -4, so this isn't ordered correctly.

0, -4, -1, 6, 11, 14

-4 is lesser than 0, so this isn't ordered correctly.

-10, -7, -4, 0, 1, 9

Each number in this sequence is ordered correctly.

0, -3, -7, 1, 4, 7

-3 is lesser than 0, so this isn't ordered correctly.

The group of numbers that is correctly ordered is -10, -7, -4, 0, 1, 9.

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The explicit formula is an= 4.2* (0.85)^{n-1}.

The correct option is (b)

<h3>What is explicit formula?</h3>

The explicit formula for a geometric sequence is of the form an = a1r-1, where r is the common ratio.

Given sequence:  {4.2, 3.57, 3.0345, 2.5793, …}

General formula a_n= ar^{n-1}

an=4.2* (17/20)^{n-1}

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brainly.com/question/18069156

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a. original \bar{x} =70,000 =\frac{\sum x}{n}

\therefore \sum x = 70,000 \times 10 =700,000

modified \sum x = 700,000 - 100,000 + 1,000,000 = 1,600,000

modified \bar{x} = \frac{1,600,000}{10} =160,000

b. Median remains the same.

c. original s^{2}=(20,000)^{2}=400000000=\frac{1}{n-1}[\sum x^{2}- n\bar{x}^{2}]

\therefore orig. \sum x^{2} = (10-1)(20000)^{2} + 10(70000)^{2}

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mod. \sum x^{2} = 52600000000-100000^{2} +1000000^{2}

                                  =1.0426E+12

mod.s^{2} = \frac{1}{n-1} [\sum x^{2} - n\bar{x}^{2}]

                       =\frac{1}{9} [1.0426E+12 - 10 \times 160000^{2}]

                       =87400000000

modified s= \sqrt{s^{2}} =\sqrt{87400000000} =295634.91            


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