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3 years ago

The graph below shows the eye color for a classroom of students.

Mathematics

1 answer:

kumpel [21]3 years ago

4 0

Answer:

it is Greater than because blue and green combined equal to 12

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The length of a rectangle is 8 inches longer than the width. If the area is 295 square inches, find the

sladkih [1.3K]

Answer:

L=21.6in and w=13.6

Step-by-step explanation:

If the length is 8 inches longer than its width, we can write the width as "w" and the length as the width w+ 8

Area is (width)(Length) = (width)(width+8)

----------

| |

| | L = w+8

| |

-----------

A = (w+8)(w)

A = w2 + 8w = 295.

(Problem states that area is 295 sq in)

Need to solve this quadratic equation

w2 +8y - 295 = 0

Factor:

(w - 13.64) (w + 21.64) = 0

w - 13.64 = 0. or. w + 21.64= 0

Solve these and get

w = 13.64. or. w = -21.64

Only one that makes sense in real life is the poitive one.

So the dimensions are

Width = 13.64 inches

Length = 21.64 inches

7 0

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

Let f(x)=5/x and g(x) =2x^2+5x. What two numbers are not in the domain of F o G?

expeople1 [14]

F(x)=5/x
g(x)=2(x^2)+5x
f(x) has a domain of all real numbers excluding zero
g(x) has a domain of all real numbers
fog(x)=5/(2(x^2)+5x)
fog(x)=5/(x(2x+5))
fog(x) has a domain that excludes both zero and -5/2

6 0

2 years ago

PLEASE SOMEONE HELP ME PLEASE :(

creativ13 [48]

Answer:

1 and 5= corresponding angles.

4 and 2=vertical opposite angle

3 and 6=co interior angle.

5 and 8=linear pair.

5 and 3= alternate interior angle

2 and 8=alternate exterior angle.

3 0