All categories

3 years ago

What is the difference between the profits Mr.Brown's store earned in the first quarter and the third quarter?

Mathematics

2 answers:

Paladinen [302]3 years ago

8 0

The answer is A, because $9,841.28 minus $7,429.84 is <span>$2,411.44</span>

DanielleElmas [232]3 years ago

6 0

Answer:

Option a. $2,411.44

Step-by-step explanation:

In the table the profit of Mr. Brown's store in all quarters of 2012 was given.

In Quarter 1 Profit was = $9,841.28

In Quarter 2 profit was = $8,957.67

In Quarter 3 profit was = $7,429.84

In Quarter 4 profit was = $11,095.67

To find out the difference between the profits earned in the first quarter and the third quarter, we subtract it.

Difference between profits = 9,841.28 - 7,429.84 = $2,411.44

Answer would be Option a. $2,411.44

You might be interested in

What sum will amount to rupees 4000 in 3 years at 6% p.a. compound interest

natali 33 [55]

Answer:

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{purple}{Given:}}}}}}}\end{gathered}$

⇢ Principle = Rs.4000
⇢ Rate = 6%
⇢ Time = 3 year

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{purple}{To Find:}}}}}}}\end{gathered}$

⇢ Amount

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{purple}{Using Formula:}}}}}}}\end{gathered}$

${\dag{\underline{\boxed{\sf{Amount ={P{\bigg(1 + \dfrac{R}{100}{\bigg)}^{T}}}}}}}}$

$\dag{\underline{\boxed{\sf{Compound \: Interest = Amount- Principle }}}}$

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{purple}{Solution:}}}}}}}\end{gathered}$

${\bigstar \:{\underline{\pmb{\frak{\red{Firstly,Finding \: the \: Amount }}}}}}$

$\quad {:\implies{\sf{Amount = \bf{P{\bigg(1 + \dfrac{R}{100}{\bigg)}^{T}}}}}}$

Substituting the values

$\quad {:\implies{\sf{Amount = \bf{4000{\bigg(1 + \dfrac{6}{100}{\bigg)}^{3}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4000{\bigg(1 \times 100 + \dfrac{6}{100}{\bigg)}^{3}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4000{\bigg( \dfrac{100 + 6}{100}{\bigg)}^{3}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4000{\bigg( \dfrac{106}{100}{\bigg)}^{3}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4000{\bigg({\cancel{\dfrac{106}{100}}{\bigg)}}^{3}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4000{\bigg( \dfrac{53}{50}{\bigg)}^{3}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4000{\bigg( \dfrac{53}{50} \times \dfrac{53}{50} \times \dfrac{53}{50}{\bigg)}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4000{\bigg( \dfrac{148877}{125000}{\bigg)}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4000 \times \dfrac{148877}{125000}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4{\cancel{000}} \times \dfrac{148877}{125{\cancel{000}}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{\dfrac{148877 \times 4}{125}}}}}$

$\quad {:\implies{\sf{Amount = \bf{\dfrac{595508}{125}}}}}$

$\quad {:\implies{\sf{Amount = \bf{\cancel{\dfrac{595508}{125}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4764.064}}}}$

$\begin{gathered} \dag{\boxed{\textsf{\textbf{\underline{\color{green}{Amount = {Rs.4764.064}}}}}}}\end{gathered}$

Hence, The Amount is Rs.4764.064

$\begin{gathered}\end{gathered}$

${\bigstar \:{\underline{\pmb{\frak{\red{ Now,Finding \: The \: Compound \: Interest }}}}}}$

$\quad{: \implies{\sf{Compound \: Interest = \bf{Amount- Principle }}}}$

Substituting the values

$\quad{: \implies{\sf{Compound \: Interest = \bf{4764.064- 4000 }}}}$

$\quad{: \implies{\sf{Compound \: Interest =\bf{764.064}}}}$

$\begin{gathered} \dag{\boxed{\textsf{\textbf{\underline{\color{green}{Compound Interest = Rs.764.064}}}}}}\end{gathered}$

Henceforth,The Compound Interest is Rs.764064

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{purple}{Learn More:}}}}}}}\end{gathered}$

$\begin{gathered}\begin{gathered}\begin{gathered} \dag \: \underline{\bf{More \: Useful \: Formula}}\\ {\boxed{\begin{array}{cc}\dashrightarrow {\sf{Amount = Principle + Interest}} \\ \\ \dashrightarrow \sf{ P=Amount - Interest }\\ \\ \dashrightarrow \sf{ S.I = \dfrac{P \times R \times T}{100}} \\ \\ \dashrightarrow \sf{P = \dfrac{Interest \times 100 }{Time \times Rate}} \\ \\ \dashrightarrow \sf{P = \dfrac{Amount\times 100 }{100 + (Time \times Rate)}} \\ \end{array}}}\end{gathered}\end{gathered}\end{gathered}$

8 0

3 years ago

In the xy-plane, the equations x + 2y = 10 and

konstantin123 [22]

By solving given equations, the value of c is 30.

Given two equations

x + 2y = 10 and

3x + 6y = c

These lines represents the same line for some constant c.

Value of c:

x + 2y = 10-------------(1)

3x + 6y = c-------------(2)

Dividing equation (2) by 3

$\frac{3x}{3}+ \frac{6y}{3} =\frac{c}{3}$

After solving the above equation, we get

x + 2y = c/3-----------(3)

Remember that a line is written as ax + by = c, in our case, both lines have a =1 and b = 2. Therefore, in orther that the two lines are equal, we need that, 10 = c/3

c = 10 × 3 = 30

c = 30

Therefore,

The value of c is 30.

Find out more information about equations here

brainly.com/question/21952977

#SPJ4

7 0

2 years ago

Which of the following is NOT a topic that you will learn about with matrices in this lesson? Addition Multiplication Interpreta

KatRina [158]

Addition

No need for stepppls it

6 0

3 years ago

Read 2 more answers

X÷-1.2= -0.3 i need help with this asap pls and thank you

Aleks [24]

0.3×1.2

=0.36

−0.3,1.2→Negative

−0.36

3 0

2 years ago

The figure is to be rotated 1/2 (180 degrees) turn about point P.

sesenic [268]

The answer is C. 8 units what you need to do is count how much from the starting point P is 4 units from A. simply double it by 2 to the new point to get 8 units.<span />

3 0

3 years ago