Simplify : $[0.9 - {2.3 - 3.2 - 3.2 - (7.1 - 5.4 - 3.5)}]$

0.18

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1.8

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2.6

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Solution

The correct option is C 0
$[0.9 - {2.3 - 3.2 - (7.1 - 5.4 - 3.5)}]$

= $[0.9 - {2.3 - 3.2 - (7.1 - 8.9)}]$

= $[0.9 - {2.3 - 3.2 - (- 1.8)}]$

= $[0.9 - {2.3 - 3.2 + 1.8}]$

= $[0.9 - {4.1 - 3.2}]$

= $[0.9 - 0.9] = 0$

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Similar questions

Q. A company manufactures car batteries of a particular type. The life (in years) of 40 batteries were recorded as follows:

2.6 3.0 3.7 3.2 2.2 4.1 3.5 4.5

3.5 2.3 3.2 3.4 3.8 3.2 4.6 3.7

2.5 4.4 3.4 3.3 2.9 3.0 4.3 2.8

3.5 3.2 3.9 3.2 3.2 3.1 3.7 3.4

4.6 3.8 3.2 2.6 3.5 4.2 2.9 3.6

Construct a grouped frequency distribution table with exclusive classes for this data, using class intervals of size

0.5

starting from the interval

2 - 2.5

Q. A company manufactures car batteries of a particular type. The lives (in years) of 40 such batteries were recorded as follows:

\begin{matrix} 2.6 & 3.0 & 3.7 & 3.2 & 2.2 & 4.1 & 3.5 & 4.5 3.5 & 2.3 & 3.2 & 3.4 & 3.8 & 3.2 & 4.6 & 3.7 2.5 & 4.4 & 3.4 & 3.3 & 2.9 & 3.0 & 4.3 & 2.8 3.5 & 3.2 & 3.9 & 3.2 & 3.2 & 3.1 & 3.7 & 3.4 4.6 & 3.8 & 3.2 & 2.6 & 3.5 & 4.2 & 2.9 & 3.6 \end{matrix}

Construct a grouped frequency distribution table for this data, using class intervals of size

0.5

starting from the interval

2 - 2.5

Question 9
A company manufactures car batteries of a particular type. The lives (in years) of 40 such batteries were recorded as follows:
$\begin{matrix} 2.6 & 3.0 & 3.7 & 3.2 & 2.2 & 4.1 & 3.5 & 4.5 3.5 & 2.3 & 3.2 & 3.4 & 3.8 & 3.2 & 4.6 & 3.7 2.5 & 4.4 & 3.4 & 3.3 & 2.9 & 3.0 & 4.3 & 2.8 3.5 & 3.2 & 3.9 & 3.2 & 3.2 & 3.1 & 3.7 & 3.4 4.6 & 3.8 & 3.2 & 2.6 & 3.5 & 4.2 & 2.9 & 3.6 \end{matrix}$
Construct a grouped frequency distribution table for this data, using class intervals of size 0.5 starting from the intervals 2−2.5.

Question 9
A company manufactures car batteries of a particular type. The lives (in years) of 40 such batteries were recorded as follows:
$\begin{matrix} 2.6 & 3.0 & 3.7 & 3.2 & 2.2 & 4.1 & 3.5 & 4.5 3.5 & 2.3 & 3.2 & 3.4 & 3.8 & 3.2 & 4.6 & 3.7 2.5 & 4.4 & 3.4 & 3.3 & 2.9 & 3.0 & 4.3 & 2.8 3.5 & 3.2 & 3.9 & 3.2 & 3.2 & 3.1 & 3.7 & 3.4 4.6 & 3.8 & 3.2 & 2.6 & 3.5 & 4.2 & 2.9 & 3.6 \end{matrix}$
Construct a grouped frequency distribution table for this data, using class intervals of size 0.5 starting from the intervals 2−2.5.

A company manufactures car batteries of a particular type. The lives (in years) of 40 such batteries were recorded as follows:
$\begin{matrix} 2.6 & 3.0 & 3.7 & 3.2 & 2.2 & 4.1 & 3.5 & 4.5 3.5 & 2.3 & 3.2 & 3.4 & 3.8 & 3.2 & 4.6 & 3.7 2.5 & 4.4 & 3.4 & 3.3 & 2.9 & 3.0 & 4.3 & 2.8 3.5 & 3.2 & 3.9 & 3.2 & 3.2 & 3.1 & 3.7 & 3.4 4.6 & 3.8 & 3.2 & 2.6 & 3.5 & 4.2 & 2.9 & 3.6 \end{matrix}$

Construct a grouped frequency distribution table for this data, using class intervals of size 0.5 starting from the intervals 2−2.5. Calculate the number of batteries having a life of 3.5-4.0 years.

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Simplify : [0.9−{2.3−3.2−3.2−(7.1−5.4−3.5)}]

The correct option is C 0[0.9−{2.3−3.2−(7.1−5.4−3.5)}]=[0.9−{2.3−3.2−(7.1−8.9)}]= [0.9−{2.3−3.2−(−1.8)}]= [0.9−{2.3−3.2+1.8}]= [0.9−{4.1−3.2}]= [0.9−0.9]=0

Simplify : $[0.9 - {2.3 - 3.2 - 3.2 - (7.1 - 5.4 - 3.5)}]$

The correct option is C 0
$[0.9 - {2.3 - 3.2 - (7.1 - 5.4 - 3.5)}]$

= $[0.9 - {2.3 - 3.2 - (7.1 - 8.9)}]$

= $[0.9 - {2.3 - 3.2 - (- 1.8)}]$

= $[0.9 - {2.3 - 3.2 + 1.8}]$

= $[0.9 - {4.1 - 3.2}]$

= $[0.9 - 0.9] = 0$