3-7-3-7-2-3-2-3-2-3-7-2-3-4-6-4-6-3-4-3-4-

The correct option is A 33/4 Given exp = #160; ( 3.7+2.3 )^2/ ( 4.6+3.4 ) ( 4.6 3.4 ) #160; #160; #160; #160; #160; #160; #1 ...

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Byju's Answer

Standard VII

Mathematics

Powers and Exponents

3.7×3.7+2.3×2...

Question

$\frac{3.7 \times 3.7 + 2.3 \times 2.3 + 2 \times 3.7 \times 2.3}{4.6 \times 4.6 - 3.4 \times 3.4} =$

3 \frac{3}{4}

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4 \frac{3}{4}

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3 \frac{1}{4}

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3 \frac{1}{2}

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Solution

The correct option is A $3 \frac{3}{4}$
Given exp = $\frac{{(3.7 + 2.3)}^{2}}{(4.6 + 3.4) (4.6 - 3.4)}$

= $\frac{6^{2}}{8 \times 1.2} = \frac{36}{9.6} = \frac{360}{96} = \frac{15}{4} = 3 \frac{3}{4}$

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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

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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}

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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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3.7×3.7+2.3×2.3+2×3.7×2.34.6×4.6−3.4×3.4=

The correct option is A 334Given exp = (3.7+2.3)2(4.6+3.4)(4.6−3.4) = 628×1.2=369.6=36096=154=334

$\frac{3.7 \times 3.7 + 2.3 \times 2.3 + 2 \times 3.7 \times 2.3}{4.6 \times 4.6 - 3.4 \times 3.4} =$

The correct option is A $3 \frac{3}{4}$
Given exp = $\frac{{(3.7 + 2.3)}^{2}}{(4.6 + 3.4) (4.6 - 3.4)}$

= $\frac{6^{2}}{8 \times 1.2} = \frac{36}{9.6} = \frac{360}{96} = \frac{15}{4} = 3 \frac{3}{4}$