All categories

2 years ago

Which story problem could be solved using the division problem 30 ÷ 5 = 6?

Mathematics

1 answer:

Lynna [10]2 years ago

8 0

Answer:

"There are 30 chairs in 5 rows. How many chairs are in each row?" Would be the correct story problem.

You might be interested in

How to divide integers

Alona [7]

I got this I hope this is helpful. To multiply or divide signed integers, always multiply or divide the absolute values and use these rules to determine the sign of the answer:
The product of two positive integers or two negative integers is positive.
The product of a positive integer and a negative integer is negative.

8 0

2 years ago

Read 2 more answers

2x+4=18 <br> what does 'x' equal?

astra-53 [7]

Answer:

Step-by-step explanation:

18-4

=14

÷2

=7.

hoped this helped

5 0

2 years ago

(8+2b)+(-4+6x)+(6+9m)

ANTONII [103]

Answer:

2 b + 9 m + 6 x + 10

Step-by-step explanation:

2 b + 4 + 6 x + 6 +9 m

2 b + 6 x + 10 + 9m

2 b + 6 x + 9 m + 10

= 2 b + 9 m + 6 x + 10

3 0

2 years ago

Read 2 more answers

39. State the equation for the line described below.

devlian [24]

Answer:

Step-by-step explanation:

39). y = $-\frac{7}{8}$ x + 3

40). m = $\frac{y_{2} -y_{1} }{x_{2} -x_{1} }$

y - $y_{1}$ = m( x - $x_{1}$ )

(- 7, 5)

( - 12, 4)

m = ( 4 - 5 ) / ( - 12 + 7) = 1/5

y - 5 = (1/5)(x + 7) ⇒ y = $\frac{1}{5}$ x + 6.4

8 0

1 year 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

2 years ago