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3 years ago

11

Julia shipped 6 packages. The lightest one weighted 24 pounds and the heaviest one weighed 67 pounds which is a reasonable range

for the total weigh of the 6 packages?

1 answer:

oee [108]3 years ago

7 0

43 pounds is a reasonable range for the total weigh of 6 packages.

Step-by-step explanation:

Given,

Number of packages shipped = 6

Weight of lightest = 24 pounds

Weight of heaviest = 67 pounds

Let,

a, b, c, d be the weight of other four packages.

24, a, b, c, d, 67 (written in lighter to heavier weight order)

Range is defined as the difference between the lowest and highest values.

Range = Highest weight - Lowest weight

Range = 67 - 24 = 43 pounds

43 pounds is a reasonable range for the total weigh of 6 packages.

Keywords: range, difference

Learn more about range at:

#LearnwithBrainly

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Find the sum of the first 20 terms of an arithmetic progression of which the third term is 55 and the last term is -98

ryzh [129]

The sum of first 20 arithmetic series $S_{20}=\frac{-3475}{16}$

Given:

Arithmetic series for 3rd term is 55

Arithmetic series for 7th term is -98

To find:

The sum of first 20 Arithmetic series

<u>Step by Step Explanation: </u>

Solution:

Formula for calculating arithmetic series

Arithmetic series=a+(n-1) d

Arithmetic series for 3rd term $a_{3}=a_{1}+(3-1) d$

$a_{1}+2 d=55$

Arithmetic series for 19th term is

$a_{19}=a_{1}+(19-1) d=-98$

$a_{19}+18 d=-98$

Subtracting equation 2 from 1

$\left[a_{19}+18 d=-98\right]+\left[a_{1}+2 d=55\right]$

16d=-98-55

16d=-153

$d=\frac{-153}{16}$

Also we know $a_{1}+2 d=55$

$a_{1}+2(-153 / 16)=55$

$a_{1}+(-153 / 8)=55$

$a_{1}=55+(153 / 8)$

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First 20 terms of an AP

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

$a_{20}=553 / 8+19(-153 / 16)$

$a_{20}=553 / 8+19(-153 / 16)$

$a_{20}=\{553 * 2 / 8 * 2\}-2907 / 16$

$a_{20}=[1106 / 16]-[2907 / 16]$

$a_{20}=-1801 / 16$

Sum of 20 Arithmetic series is

$S_{n}=n\left(a_{1}+a_{n}\right) / 2$

Substitute the known values in the above equation we get

$S_{20}=\left[\frac{20\left(\left(\frac{558}{8}\right)+\left(\frac{-1801}{16}\right)\right)}{2}\right]$

$S_{20}=\left[\frac{\left.20\left(\frac{1106}{16}\right)+\left(\frac{-1801}{16}\right)\right)}{2}\right]$

$S_{20}=10 \frac{(-695 / 16)}{2}$

$S_{20}=5\left[\frac{-695}{16}\right]$

$S_{20}=\frac{-3475}{16}$

Result:

Thus the sum of first 20 terms in an arithmetic series is $S_{20}=\frac{-3475}{16}$

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