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MAXImum [283]
2 years ago
8

John needs 20 grams of 54% acid solution for his science project. His school's science lab has bottles of 30% solution and bottl

es of 60% solution. Hiw much of each should john use?
Mathematics
2 answers:
anzhelika [568]2 years ago
8 0

Answer:

4 grams of 30% solution and 16 grams of 60% solution

Step-by-step explanation:

Paul [167]2 years ago
7 0

Answer:

John should use:

4 grams of the 30% solution and 16 grams of the 60% solution

Step-by-step explanation:

Let the number of grams of the 30% solution = x

Let the number of grams of the 60% solution = y

John needs 20 grams of 54% acid solution for his science project.

Hence,

x + y = 20 grams..... Equation 1

x = 20 - y

His school's science lab has bottles of 30% solution and bottles of 60% solution.

30% × x + 60% × y = 54% × 20

0.3x + 0.6y = 10.8......Equation 2

We substitute 20 - y for x in Equation 2

0.3(20 - y) + 0.6y = 10.8

6 - 0.3y + 0.6y = 10.8

- 0.3y + 0.6y = 10.8 - 6

0.3y = 4.8

y = 4.8/3

y = 16 grams

x = 20 - y

x = 20 - 16

x = 4 grams

Therefore, John should use:

4 grams of the 30% solution and 16 grams of the 60% solution

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Helppp me plsssssssss<br><br>​
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The class 35 - 40 has maximum frequency. So, it is the modal class.

From the given data,

  • \sf \:\:\:\:\:\:\:\:\:\:x_{k}=35
  • \sf \:\:\:\:\:\:\:\:\:\:f_{k}=50
  • \sf \:\:\:\:\:\:\:\:\:\:f_{k-1}=34
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{\bf \:\: {By\:using\:the\: formula}} \\ \\

\:\dag\:{\small{\underline{\boxed{\sf {Mode,\:M_{o} =\sf\red{x_k + {\bigg(h \times \: \dfrac{ ( f_k - f_{k-1})}{ (2f_k - f_{k - 1} - f_{k +1})}\bigg)}}}}}}} \\ \\

\sf \:\:\:\:\:\:\:\:\:= 35+ {\bigg(5 \times \dfrac{(50 - 34)}{ ( 2 \times 50 - 34 - 42)}\bigg)} \\ \\

\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= 35 +{\bigg(5 \times \dfrac{16}{24}\bigg)} \\ \\

\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= {\bigg(35+\dfrac{10}{3}\bigg)} \\ \\

\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:(35 + 3.33) =.38.33 \\ \\

\:\:\sf {Hence,}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\ \large{\underline{\mathcal{\gray{ mode\:=\:38.33}}}} \\ \\

{\large{\frak{\pmb{\underline{Additional\: information }}}}}

MODE

  • Most precisely, mode is that value of the variable at which the concentration of the data is maximum.

MODAL CLASS

  • In a frequency distribution the class having maximum frequency is called the modal class.

{\bf{\underline{Formula\:for\: calculating\:mode:}}} \\

{\underline{\boxed{\sf {Mode,\:M_{o} =\sf\red{x_k + {\bigg(h \times \: \dfrac{ ( f_k - f_{k-1})}{ (2f_k - f_{k - 1} - f_{k +1})}\bigg)}}}}}} \\ \\

Where,

\sf \small\pink{ \bigstar} \: x_{k}= lower\:limit\:of\:the\:modal\:class\:interval.

\small \blue{ \bigstar}\sf \: f_{k}=frequency\:of\:the\:modal\:class

\sf \small\orange{ \bigstar}\: f_{k-1}=frequency\:of\:the\:class\: preceding\:the\;modal\:class

\sf \small\green{ \bigstar}\: f_{k+1}=frequency\:of\:the\:class\: succeeding\:the\;modal\:class

\small \purple{ \bigstar}\sf \: h= width \:of\:the\:class\:interval

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