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3 years ago

Add the linear expressions. 2x + 1 and x + 2

Mathematics

2 answers:

sdas [7]3 years ago

6 0

Answer:

3(x + 1)

Step-by-step explanation:

By adding

(2x + 1) + (x+2)

2x +1 +x +2

3x + 3 or 3(x + 1)

solong [7]3 years ago

6 0

2x + 1 + x+2
2x +x +1+2
3x +3

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3 years ago

Paolo is using his grandmother's breakfast bar recipe. He always doubles

Papessa [141]

Answer:

$6\frac{1}{3}$ = $2\frac{1}{2}$ * 2 + $\frac{x}{2}$

Step-by-step explanation:

lets start with what we know.

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2 years ago

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2 years ago

The winning time for the middle school 4 person 100 meter relay 62.59 seconds. Suppose that each runner ran exactly the same amo

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2 years ago

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Oliga [24]

Answer:

The class 35 - 40 has maximum frequency. So, it is the modal class.

From the given data,

$\sf \:\:\:\:\:\:\:\:\:\:x_{k}=35$
$\sf \:\:\:\:\:\:\:\:\:\:f_{k}=50$
$\sf \:\:\:\:\:\:\:\:\:\:f_{k-1}=34$
$\sf \:\:\:\:\:\:\:\:\:\:f_{k+1}=42$
$\sf \:\:\:\:\:\:\:\:\:\:h=5$

${\bf \:\: {By\:using\:the\: formula}} \\ \\$

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

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

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

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

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

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

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

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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3 years ago