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$\text{32}+\text{68}+\text{29}+\text{73}$
Notice two things:
The sum may be estimated by
$\begin{array}{ccc}\hfill \left(2\cdot 30\right)+\left(2\cdot \text{70}\right)& =& 6+140\hfill \\ & =& \text{200}\hfill \end{array}$
In fact, $\text{32}+\text{68}+\text{29}+\text{73}=\text{202}$ .
Estimate each sum. Results may vary.
$\text{27}+\text{48}+\text{31}+\text{52}$ .
27 and 31 cluster near 30. Their sum is about $\text{2}\cdot \text{30}=\text{60}$ .
48 and 52 cluster near 50. Their sum is about $\text{2}\cdot \text{50}=\text{100}$ .
Thus, $\text{27}+\text{48}+\text{31}+5\text{2}$ is about $\begin{array}{ccc}(2\cdot 30)+(2\cdot 50)& =& 60+100\hfill \\ & =& 160\hfill \end{array}$
In fact, $\text{27}+\text{48}+\text{31}+\text{52}=\text{158}$ .
$\text{88}+\text{21}+\text{19}+\text{91}$ .
88 and 91 cluster near 90. Their sum is about $\text{2}\cdot \text{90}=\text{180}$ .
21 and 19 cluster near 20. Their sum is about $\text{2}\cdot \text{20}=\text{40}$ .
Thus, $\text{88}+\text{21}+\text{19}+\text{91}$ is about $\begin{array}{ccc}(2\cdot 90)+(2\cdot 20)& =& 180+40\hfill \\ & =& 220\hfill \end{array}$
In fact, $\text{88}+\text{21}+\text{19}+\text{91}=\text{219}$ .
$\text{17}+\text{21}+\text{48}+\text{18}$ .
17, 21, and 18 cluster near 20. Their sum is about $\text{3}\cdot \text{20}=\text{60}$ .
48 is about 50.
Thus, $\text{17}+\text{21}+\text{48}+\text{18}$ is about $\begin{array}{ccc}(3\cdot 20)+50& =& 60+50\hfill \\ & =& 110\hfill \end{array}$
In fact, $\text{17}+\text{21}+\text{48}+\text{18}=\text{104}$ .
$\text{61}+\text{48}+\text{49}+\text{57}+\text{52}$ .
61 and 57 cluster near 60. Their sum is about $\text{2}\cdot \text{60}=\text{120}$ .
48, 49, and 52 cluster near 50. Their sum is about $3\cdot \text{50}=\text{150}$ .
Thus, $\text{61}+\text{48}+\text{49}+\text{57}+\text{52}$ is about $\begin{array}{ccc}(2\cdot 60)+(3\cdot 50)& =& 120+150\hfill \\ & =& 270\hfill \end{array}$
In fact, $\text{61}+\text{48}+\text{49}+\text{57}+\text{52}=\text{267}$ .
$\text{706}+\text{321}+\text{293}+\text{684}$ .
706 and 684 cluster near 700. Their sum is about $\text{2}\cdot \text{700}=\text{1,400}$ .
321 and 293 cluster near 300. Their sum is about $\text{2}\cdot \text{300}=\text{600}$ .
Thus, $\text{706}+\text{321}+\text{293}+\text{684}$ is about $\begin{array}{ccc}(2\cdot 700)+(2\cdot 300)& =& \mathrm{1,400}+600\hfill \\ & =& \mathrm{2,000}\hfill \end{array}$
In fact, $\text{706}+\text{321}+\text{293}+\text{684}=\text{2,004}$ .
Use the clustering method to estimate each sum.
$\text{28}+\text{51}+\text{31}+\text{47}$
$(2\cdot 30)+(2\cdot 50)=60+100=160$
$\text{42}+\text{39}+\text{68}+\text{41}$
$(3\cdot 40)+70=120+70=190$
$\text{37}+\text{39}+\text{83}+\text{42}+\text{79}$
$(3\cdot 40)+(2\cdot 80)=120+160=280$
$\text{612}+\text{585}+\text{830}+\text{794}$
$(2\cdot 600)+(2\cdot 800)=\mathrm{1,200}+\mathrm{1,600}=\mathrm{2,800}$
Use the clustering method to estimate each sum. Results may vary.
$\text{28}+\text{51}+\text{31}+\text{47}$
$2\left(\text{30}\right)+2\left(\text{50}\right)=\text{160}\left(\text{157}\right)$
$\text{42}+\text{19}+\text{39}+\text{23}$
$\text{88}+\text{62}+\text{59}+\text{90}$
$2\left(\text{90}\right)+2\left(\text{60}\right)=\text{300}\left(\text{299}\right)$
$\text{76}+\text{29}+\text{33}+\text{82}$
$\text{19}+\text{23}+\text{87}+\text{21}$
$3\left(\text{20}\right)+\text{90}=\text{150}\left(\text{150}\right)$
$\text{41}+\text{28}+\text{42}+\text{37}$
$\text{89}+\text{32}+\text{89}+\text{93}$
$3\left(\text{90}\right)+\text{30}=\text{300}\left(\text{303}\right)$
$\text{73}+\text{72}+\text{27}+\text{71}$
$\text{43}+\text{62}+\text{61}+\text{55}$
$\text{40}+3\left(\text{60}\right)=\text{220}\left(\text{221}\right)$
$\text{31}+\text{77}+\text{31}+\text{27}$
$\text{57}+\text{34}+\text{28}+\text{61}+\text{62}$
$3\left(\text{60}\right)+2\left(\text{30}\right)=\text{240}\left(\text{242}\right)$
$\text{94}+\text{18}+\text{23}+\text{91}+\text{19}$
$\text{103}+\text{72}+\text{66}+\text{97}+\text{99}$
$3\left(\text{100}\right)+2\left(\text{70}\right)=\text{440}\left(\text{437}\right)$
$\text{42}+\text{121}+\text{119}+\text{124}+\text{41}$
$\text{19}+\text{24}+\text{87}+\text{23}+\text{91}+\text{93}$
$3\left(\text{20}\right)+3\left(\text{90}\right)=\text{330}\left(\text{337}\right)$
$\text{108}+\text{61}+\text{63}+\text{96}+\text{57}+\text{99}$
$\text{518}+\text{721}+\text{493}+\text{689}$
$2\left(\text{500}\right)+2\left(\text{700}\right)=\mathrm{2,}\text{400}\left(\mathrm{2,}\text{421}\right)$
$\text{981}+\text{1208}+\text{1214}+\text{1006}$
$\text{23}+\text{81}+\text{77}+\text{79}+\text{19}+\text{81}$
$2\left(\text{20}\right)+4\left(\text{80}\right)=\text{360}\left(\text{360}\right)$
$\text{94}+\text{68}+\text{66}+\text{101}+\text{106}+\text{71}+\text{110}$
( [link] ) Find the product: $\frac{2}{3}\cdot \frac{9}{\text{14}}\cdot \frac{7}{\text{12}}$ .
( [link] ) Convert 0.06 to a fraction.
$\frac{3}{\text{50}}$
( [link] ) Write the proportion in fractional form: "5 is to 8 as 25 is to 40."
( [link] ) Estimate the sum using the method of rounding: $\text{4,882}+\text{2,704}$ .
$\mathrm{4,}\text{900}+\mathrm{2,}\text{700}=\mathrm{7,}\text{600}(\mathrm{7,}\text{586})$
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