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Kay [80]
3 years ago
12

A used-car dealer has a vehicle on the lot with a sticker price of $5999. If the

Mathematics
1 answer:
Digiron [165]3 years ago
3 0

Answer:

$4999

Step-by-step explanation:

Let the cost of the car be x

<u>Then considering 20% markup, the sticker price is:</u>

  • x + 20% = 5999

<u>Since 20% of x is 20/100x = 0.2x,</u>

  • x+ 0.2x = 5999
  • 1.2x = 5999
  • x = 5999/1.2
  • x ≈ $4999

<u>So dealer paid</u> $4999 for the car

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1809<br>×308<br>_____<br><br>______​
klio [65]

Answer:

557152

Step-by-step explanation:

You first multiply the 8 in 308 to the 9 in 1809, then the 0, then the 9, then finally the 1 in 1809. For the tens place, don't forget to put a zero under the ones place because that's not where you start. So for 0 everything is 0 so just put a line of 5 zeros in the second row. Lastly, the 3 multiplies to 9, 0, 8, and the 1 in 1809. In the 3rd row, put 2 zeros because you're multiplying to get the hundreds place. Line and add them all up to get 557152. Hope this helps!

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3 years ago
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Surfside Bike Rental Shop charges a $20 fixed fee plus $5 per hour for renting a bike. Alyssa paid $50 to rent a bike. How many
Leviafan [203]

Answer:

D.

Step-by-step explanation:

You can set up and equation to solve for any cost like so:

5x + 20 = c

5x is the price for each hour

20 is the fixed fee

c is the total cost

Now let's use the price given to solve:

5x + 20 = 50

Subtract 20 from each side:

5x = 30

Divide each side by 5:

x = 6

5 0
3 years ago
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Find the value of each of these quantities. <br> C(5,3)<br> C(8,8)<br> C(12,6)
Paul [167]
C^n_k= \frac{n!}{(n-k)!c!} &#10;\\&#10;\\C^5_3= \frac{5!}{(5-3)!3!} = \frac{5!}{2!3!} = \frac{5\times4\times3}{3\times2\times1} =10&#10;\\&#10;\\C^8_8= \frac{8!}{(8-8)!8!} = \frac{8!}{0!8!} = \frac{8!}{1\times8!}=1&#10;\\&#10;\\C^{12}_6= \frac{12!}{(12-6)!6!} = \frac{12!}{6!6!} = \frac{12\times11\times10\times9\times8\times7}{6\times5\times4\times3\times2\times1} =\frac{11\times3\times4\times7}{1} =924
7 0
3 years ago
the sum of the speeds of two trains is 716.7 miles per hour if the speed of the first train is 5.3 miles per hour faster than a
Nimfa-mama [501]
The best way to solve this is to make an algebraic expression to represent the situation~
If x is the speed of the first train and y is the speed of the second we can make 716.7 = x + y
the first train is 5.3 mph faster so x = y + 5.3
Now we can replace x with our new value to get 716.7 = y + (y+5.3)
Simplify and we get 716.7 = 2y + 5.3
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5 0
3 years ago
Please help me! 20 points! You don't have to graph btw! This is my final question!
FrozenT [24]

Your \ answer \ is.

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