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trasher [3.6K]
3 years ago
12

Alberto measures the length of a piece of wire to be 105 cm. The actual length of the wire was 120 cm. What was the percent erro

r of Alberto's measurement? Show your work. ​
Mathematics
1 answer:
natali 33 [55]3 years ago
7 0

Answer:

12.5%

Step-by-step explanation:

105 over 120 is the fraction that he got so you subtract 105 from 120 to get how much he missed. 120-105=15

He miscalculated by 15 cm out of the total of 120, 15/ 120

15/120=5/40=2.5/20 multiply the top and bottom by 5 to get the percentage out of 100. 2.5 times 5 =12.5, so he miscalculated by 12.5%

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Two similar cones have volume of 4 m and 108 m respectively. If the large one has surface area 54 m, find the surface area of th
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\displaystyle\bf\\Explanations:\\\\The~similarity~ratio~of~two~similar~cones=k\\\\k=is~the~ratio~between~2~corresponding~lengths\\\\k=\frac{R_1}{R_2}=\frac{h_1}{h_2}\\\\The~ratio~between~2~corresponding~areas~of~similar~cones=k^2\\\\\frac{Area~1}{Area~2}=k^2\\\\The~ratio~between~the~volumes~of~the~2~similar~cones=k^3\\\\\frac{Volume~1}{Volume~2}=k^3

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\displaystyle\bf\\Solving:\\\\Volume1=108~m^3\\\\Volume2=4m^3\\\\k^3=\frac{108}{4}\\\\k^3=27~~\Big|\sqrt{~}\\\\k=\sqrt[\b3]{27}\\\\k=3\\\\Area1=54~m^2\\\\\frac{Area~1}{Area~2}=k^2\\\\\frac{54}{Area~2}=3^2\\\\\frac{54}{Area~2}=9\\\\Area2=\frac{54}{9}\\\\\boxed{\bf Area2=6 m^2}

 

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