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3 years ago

Convert a 30% markup percent on cost to markup percent on selling price. (Round to the nearest hundredth percent.)

Mathematics

1 answer:

Art [367]3 years ago

4 0

Answer:

Option C

23.08% markup on selling price.

Step-by-step explanation:

Given in the question,

markup percentage on cost price = 30%

To find,

markup percentage on selling price

Markup is the ratio between the cost of a good or service and its selling price.

Let suppose that cost price percentage = 100%

As we know that,

<h3>cost% + markup% = selling%</h3><h3>100% + 30% = 130%</h3>

So percent markup selling price = 30 / 130 x 100

= 23.0769

Hence, 30% markup on cost price = 23.0769% markup on selling price.

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Step-by-step explanation:

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Let x be the ages of students in a class. Given the frequency distribution f(x) above, determine the following probabilities:.

lora16 [44]

Probabilities are used to determine the chances of events

The value of the probability P(x < 24) is 0.7895

<h3>How to calculate the probability P(x < 24)?</h3>

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$P(x < 24) = \frac{n(x < 24)}{Total}$

So, we have:

$P(x < 24) = \frac{22 + 3 + 5}{22 + 3 + 5 + 4 + 4}$

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Hence, the value of the probability P(x < 24) is 0.7895

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2 years ago

Describe the translation of the function from the parent function f(x) = x².

kkurt [141]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\

\begin{array}{rllll} 
% left side templates
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
\\\\

\end{array}\\$

$\bf \begin{array}{llll}
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\bullet \textit{ vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}
\end{array}
$

now, notice the template above... now let's see your function $\bf y=x^2+4\implies 
\begin{array}{llllll}
y=&1(&1x&+&0)^2+&4\\
&A&B&&C&D
\end{array}$

so.. what do you think was the shift then?

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3 years ago

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Vlad [161]

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$P(B|A) = \frac{P(A \cap B)}{P(A)}$