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2 years ago

Find the Mean from the line plot. Show your work and explain how you solved using First, Next, and Last.

Mathematics

1 answer:

Musya8 [376]2 years ago

8 0

Answer:

mean is 5

Step-by-step explanation:

mean is regarded as the average number of items

<h3>mean formula:</h3>

<em>(sum of terms) / (number of terms)</em>

<h3><u /></h3><h3><u>Firstly</u></h3>

the numbers in the data set are :

3,3,3,4,6,6,7,8

they are already arranged in order, so no need to rearrange.

add the numbers together to find the total :

3 +3 +3 +4 +6+ 6+7 +8

= 40

ALSO,

find how many number of terms they are --- this is 8

Last

use the mean formula to solve :

<em>(sum of terms) / (number of terms)</em>

= 40/8

= 5

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A sequence is given by rule 4n-5 generate the first 5 terms of the sequence

gtnhenbr [62]

Answer:

- 1, 3, 7, 11, 15

Step-by-step explanation:

Substitute n = 1, 2, 3, 4, 5 into the rule, that is

a₁ = 4(1) - 5 = 4 - 5 = - 1

a₂ = 4(2) - 5 = 8 - 5 = 3

a₃ = 4(3) - 5 = 12 - 5 = 7

a₄ = 4(4) - 5 = 16 - 5 = 11

a₅ = 4(5) - 5 = 20 - 5 = 15

The first 5 terms are - 1, 3, 7, 11, 15

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Step-by-step explanation:

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2 years ago

40, 10, 5/2 5/8... (fractions) is it arithmetic or geometric and what are the next two terms PLZ HELP DUE IN 10 MINS

patriot [66]

Answer:

Please check the explanation.

Step-by-step explanation:

Given the sequence

$40,\:10,\:\frac{5}{2},\:\frac{5}{8}$

A geometric sequence has a constant ratio 'r' and is defined by

$\:a_n=a_0\cdot r^{n-1}$

Computing the ratios of all the adjacent terms

$\frac{10}{40}=\frac{1}{4},\:\quad \frac{\frac{5}{2}}{10}=\frac{1}{4},\:\quad \frac{\frac{5}{8}}{\frac{5}{2}}=\frac{1}{4}$

The ratio of all the adjacent terms is the same and equal to

$r=\frac{1}{4}$

Thus, the given sequence is a geometric sequence.

As the first element of the sequence is

$a_1=40$

Therefore, the nth term is calculated as

$\:a_n=a_0\cdot r^{n-1}$

$a_n=40\left(\frac{1}{4}\right)^{n-1}$

Put n = 5 to find the next term

$a_5=40\left(\frac{1}{4}\right)^{5-1}$

$a_5=40\cdot \frac{1}{4^4}$

$a_5=\frac{40}{4^4}$

$=\frac{2^3\cdot \:5}{2^8}$

$a_5=\frac{5}{2^5}$

$a_5=\frac{5}{32}$

now, Put n = 6 to find the 6th term

$a_6=40\left(\frac{1}{4}\right)^{6-1}$

$a_6=40\cdot \frac{1}{4^5}$

$a_6=\frac{40}{4^5}$

$=\frac{2^3\cdot \:5}{2^{10}}$

$a_6=\frac{5}{2^7}$

$a_6=\frac{5}{128}$

Thus, the next two terms of the sequence 40, 10, 5/2, 5/8... is:

$a_5=\frac{5}{32}$

$a_6=\frac{5}{128}$

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