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3 years ago

A department store buys 200 shirts at a cost of $5,400 and sells them at a selling

Mathematics

2 answers:

riadik2000 [5.3K]3 years ago

7 0

Answer:

11.11%

so mark me as brainlist

jekas [21]3 years ago

3 0

Number of shirts a department store bought = 200

Cost price of 200 shirts = $5400

Cost price of 1 shirt :

$=\tt \frac{5400}{200}$

$\color{plum} =\tt \$27$

Thus, the cost price of one shirt = $27

Price at which the department store sold each shirt = $30

Which means :

▪︎Cost price of shirt = $27

▪︎Selling price of shrit = $30

Since selling price is greater than cost price, there was a markup in the price of the shirts.

We know that :

$\color{hotpink}\tt \: markup \: percentage = \color{plum} \frac{selling \: price - cost \: price}{cost \: price } \times 100$

Markup percentage in the price of the shirts :

$= \tt \frac{30 - 27}{27} \times 100$

$= \tt \frac{3}{27} \times 100$

$= \tt \frac{3 \times 100}{27}$

$= \tt \frac{300}{27}$

$\color{plum} = \tt11.1 \%$

Thus, the percentage of markup = $\tt 11.1 \%$

Therefore, the percentage of markup in the shirt's price = $\color{hotpink} \tt11.1 \%$

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Mr.Jimerson earns $24 per hour working. He qualifies for a 15% raise in salary. What is his salary after his raise?

miv72 [106K]

(sorry if this is wrong but hope this helps)

ANSWER: $27.60

STEP BY STEP:

You have to multiply 24 by 15%. Start by changing 15% into a decimal of .15. Then multiply 24 by .15 to get an answer of $3.60. This is how much more Mr. Jimerson will earn. Then add the original amount he earned by the amount the is now gaining. This gets you $27.60.

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3 years ago

Marisol is making a rectangular wooden frame. She wants the length of the frame to be no more than 12 inches. She has less than

NISA [10]

Answer:

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Step-by-step explanation:

l ≤ 12

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3 years ago

Read 2 more answers

Which of the following graphs shows the solution set for the inequality below? 3|x + 1| < 9

Bas_tet [7]

Step-by-step explanation:

The absolute value function is a well known piecewise function (a function defined by multiple subfunctions) that is described mathematically as

$f(x) \ = \ |x| \ = \ \left\{\left\begin{array}{ccc}x, \ \text{if} \ x \ \geq \ 0 \\ \\ -x, \ \text{if} \ x \ < \ 0\end{array}\right\}$ .

This definition of the absolute function can be explained geometrically to be similar to the straight line $\textbf{\textit{y}} \ = \ \textbf{\textit{x}}$ , however, when the value of x is negative, the range of the function remains positive. In other words, the segment of the line $\textbf{\textit{y}} \ = \ \textbf{\textit{x}}$ where $\textbf{\textit{x}} \ < \ 0$ (shown as the orange dotted line), the segment of the line is reflected across the x-axis.

First, we simplify the expression.

$3\left|x \ + \ 1 \right| \ < \ 9 \\ \\ \\\-\hspace{0.2cm} \left|x \ + \ 1 \right| \ < \ 3$ .

We, now, can simply visualise the straight line, $y \ = \ x \ + \ 1$ , as a line having its y-intercept at the point $(0, \ 1)$ and its x-intercept at the point $(-1, \ 0)$ . Then, imagine that the segment of the line where $x \ < \ 0$ to be reflected along the x-axis, and you get the graph of the absolute function $y \ = \ \left|x \ + \ 1 \right|$ .

Consider the inequality

$\left|x \ + \ 1 \right| \ < \ 3$ ,

this statement can actually be conceptualise as the question

$``\text{For what \textbf{values of \textit{x}} will the absolute function \textbf{be less than 3}}"$ .

Algebraically, we can solve this inequality by breaking the function into two different subfunctions (according to the definition above).

Case 1 (when $x \ \geq \ 0$ )

$x \ + \ 1 \ < \ 3 \\ \\ \\ \-\hspace{0.9cm} x \ < \ 3 \ - \ 1 \\ \\ \\ \-\hspace{0.9cm} x \ < \ 2$

Case 2 (when $x \ < \ 0$ )

$-(x \ + \ 1) \ < \ 3 \\ \\ \\ \-\hspace{0.15cm} -x \ - \ 1 \ < \ 3 \\ \\ \\ \-\hspace{1cm} -x \ < \ 3 \ + \ 1 \\ \\ \\ \-\hspace{1cm} -x \ < \ 4 \\ \\ \\ \-\hspace{1.5cm} x \ > \ -4$

*remember to flip the inequality sign when multiplying or dividing by

negative numbers on both sides of the statement.

Therefore, the values of x that satisfy this inequality lie within the interval

$-4 \ < \ x \ < \ 2$ .

Similarly, on the real number line, the interval is shown below.

The use of open circles (as in the graph) indicates that the interval highlighted on the number line does not include its boundary value (-4 and 2) since the inequality is expressed as "less than", but not "less than or equal to". Contrastingly, close circles (circles that are coloured) show the inclusivity of the boundary values of the inequality.