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Andrews [41]
3 years ago
15

Find the selling price of a $20 item after a 20% markup.

Mathematics
1 answer:
nevsk [136]3 years ago
4 0

Answer:

$24

Step-by-step explanation:

For a markup, use the formula <em>a = ( 1 + p ) × w. </em>

!!! Remember, <em>a </em>is the selling price, and <em>w </em>is the original price !!!

!!! The <em>p </em>is the percent in decimal form. 20% as a decimal is 0.20 (move the decimal twice to the left) !!!

<em />

Inserting the values, it should look like <em>a = ( 1 + 0.20 ) × 20 </em>

<em>a = 1.20 × 20 </em>

<em>a = 24</em>

<em />

Hence, the new selling price is $24.

Hope this helps! :)

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Nathan has just bought a new car. He models the value, V, in dollars, of the car after t years as V(t) = 21,000(0.861)t. Based o
UkoKoshka [18]

Answer:

13.9%

Step-by-step explanation:

The value of car is modeled as:

V(t)=21,000(0.861)t

Here we can see that, each year Nathan has considered 0.861 of the previous year value or we can say that Nathan has considered 86.1% of the previous year value. So,

100-86.1=13.9

We subtract the percentage value considered from 100 to find out the percentage decrease in the value of the car.

The value of new car is decreasing by 13.9% each year.  

4 0
3 years ago
If a cup of coffee has temperature 97°C in a room where the ambient air temperature is 25°C, then, according to Newton's Law o
FinnZ [79.3K]

Answer:

The  average temperature  is T_{a} = 81.95^oC

Step-by-step explanation:

From the question we are told that

    The temperature of the coffee after time t is   T(t) =  25 + 72 e^{[-\frac{t}{45} ]}

Now the average temperature during the first 22 minutes i.e fro 0 \to  22minutes is mathematically evaluated as

              T_{a} =  \frac{1}{22-0}  \int\limits^{22}_{0} {25 +72 e^{[-\frac{t}{45} ]}} \, dx

               T_{a} = \frac{1}{22} [25 t  +  72 [\frac{e^{[-\frac{t}{45} ]}}{-\frac{1}{45} } ] ] \left| 22} \atop {0}} \right.

             T_{a} = \frac{1}{22} [25 t  - 3240e^{[-\frac{t}{45} ]} ] \left | 45} \atop {{0}} \right.

              T_{a} = \frac{1}{22} [25 (22)  - 3240e^{[-\frac{22}{45} ]}   - (- 3240e^{0} )]

            T_{a} = \frac{1}{22} [550  - 1987.12  +  3240]

          T_{a} = 81.95^oC

       

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Answer:

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