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3 years ago

59742

Mathematics

1 answer:

Sergeu [11.5K]3 years ago

8 0

Answer:

Step-by-step explanation:

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Answer ASAP pls pls

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D be the correct answer

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3 years ago

Find the slope of the line passing through the pairs of points and describe the line as rising, falling, horizontal or vertical.

Anna [14]

Answer:

$a.\ m=2,\ \text{the line is rising}\\\\b.\ m=-\dfrac{5}{4},\ \text{the line is falling}\\\\c.\ m=0,\ \text{the line is horizontal}\\\\d.\ m\ is\ unde fined,\ \text{the line is vertical}$

Step-by-step explanation:

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

m > 0, then a line is rising

m < 0, then a line is falling

m = 0, then a line is horizontal

m is undefined, then a line is vertical

(2, 1) and (4, 5)

$m=\dfrac{5-1}{4-2}=\dfrac{4}{2}=2>0\to\text{rising}$

(-1, 0) and (3, -5)

$m=\dfrac{-5-0}{3-(-1)}=\dfrac{-5}{4}=-\dfrac{5}{4}$

(2, 1) and (-3, 1)

$m=\dfrac{1-1}{-3-2}=\dfrac{0}{-5}=0\to\text{horizontal}$

(-1, 2) and (-1, -5)

$m=\dfrac{-5-2}{-1-(-1))}=\dfrac{-7}{0}\ \text{UNDEFINED}\to\text{vertical}$

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3 years ago

Help please ! solve for x and round the answer to the nearest tenth. need it asap , thank you

mafiozo [28]

Answer:

3.6

Step-by-step explanation:

11x=40

X=40/11

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3 years ago

What is the positive solution to this equation? 4x2 + 12x = 135

Svet_ta [14]

Answer:

x = 9/2 - 15/2

Step-by-step explanation:

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3 years ago

Helppp me plsssssssss<br><br>

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,

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

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

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

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

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

7 0

3 years ago