What is $220 marked up 60%​

Answers

Answer 1
Answer:

A $220 item marked up by 60% would have a final selling price of $352.

what is $220 marked up 60%

A 60% markup on $220 would be calculated as follows:

Markup = Original Price × Markup Percentage

Markup = $220 × 0.60

Markup = $132

To find the final selling price, you add the markup to the original price:

Final Price = Original Price + Markup

Final Price = $220 + $132

Final Price = $352

Therefore, a $220 item marked up by 60% would have a final selling price of $352.

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Answer 2
Answer:

Answer:

Answer would be $366.70

Step-by-step explanation:

220/60%=366.70

Hope it helps


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The equation of a circle is (x - 3)2 + (y + 2)2 = 25. The point (8, -2) is on the circle.What is the equation of the line that is tangent to the circle at (8, -2)? y = 8 x = 8 x = 3 y = 3
(-3x=2)-(3-5x-1)

Solve equation 2x^{2}−5x−7 = 0 by using the quadratic formula. Then write 2x^{2}−5x−7 in form of a(x−r_{1})(x−r_{2}).

Answers

2x^2+2x-7x-7=2x(x+1)-7(x+1)=(2x-7)(x+1)=\n \n=2(x-3.5)(x+1)\n \n2x^2-5x-7=0\ \ \ \Leftrightarrow\ \ \ 2(x-3.5)(x+1)=0\n \nx-3.5=0\ \ \vee\ \ \ x+1=0\n \n.\ \ \ \ \ \ \ x=3.5\ \ \ \ \ \ \ \ \ \ \ x=-1

3,9,6,9,27,24,27,81,78,81,243,?find out last figure to complete series

Answers

the next number is 240

A man's regular pay is $3 per hour up to 40 hours. Overtime is twice the payment for regular time If he was paid $168, how many hours overtime did he work?

Answers

Answer:

He worked 8 hours overtime.

Step-by-step explanation:

A man was paid per hour up to 40 hours = $3.00

He was paid for 40 hours = 40 × 3 = $120

He did overtime, for which he was paid twice of the regular payment = 2(3)

= $6.00

He was paid for his overtime = 168 - 120 = $48.00

His overtime payment per hour = $6.00

He did overtime = 48 ÷ 6 = 8 hours

He worked 8 hours overtime.

He worked 8 more hours overtime. Good luck with the rest of your homework..

If PQ = 21 cm and QR = 5, then what are the possible lengths for PR so that PQ,QR and PR can form a triangle? Explain your reasoning

Answers

Please see the pic, I'd solved in it.
So we are given the measurements of two sides of a Triangle:
(write down the given to be clear)
PQ=21 cm 
QR=5 cm
But one very important thing:
As Aman said, the length of PR would definitely not be greater than the sum of the other two sides/legs....
Why? Because then it gives a STRAIGHT LINE!
which means: PR will be less than that sum
PR<sum of PQ and QR
PR<21+5
PR<26
Time for Pythagoras Theorem!
We can now split it into Two ways, since we dont know if PR is just a leg or the hypotenuse of the triangle:
And therefore:
Assuming PR is a leg:
(PQ seems to be the larger number)
PQ²=PR²+QR²
21² =PR²+5²
21²-5²=PR²
√21²-5²=PR
20.396 cm=PR
≈20.4 cm
20.4≤PR<26  (which means it can be 20.4 as well, but not any lower as                        it is very obvious)
Assuming that PR is the hypotenuse:
PR²=PQ²+QR²
PR²=21²+5²
PR=√21²+5²
PR=21.587 cm≈ 21.6 cm
21.6≤PR<26
So basically, the length can be 21.6 itself or higher.


Which of the following is the solution set of the given equation?
14 + 8m = 14 - 3m - 5m

Answers

14 + 8m = 14 - 3m - 5m
14 + 8m = 14 - 8m (Subtract 3m and 5m)
14 + 16m = 14       (Add 8m to both sides)
16m = 0                 (Subtract 14 from both sides)
m = 0                     (Divide 16 from both sides)

Answer:

m= 0

Step-by-step explanation:

14 + 8m = 14 - 3m - 5m

14 + 8m = 14 - 8m     (Subtract 3m and 5m)

14 + 16m = 14       (Add 8m to both sides)

16m = 0                 (Subtract 14 from both sides)

m = 0                    (Divide 16 from both sides)

How was pre-image CDEF transformed to create image CꞌDꞌEꞌFꞌ?A. 180° clockwise rotation around point G


B. 270° counterclockwise rotation around point G


C. reflection over segment CF


D. translation 2 units to the right

Answers

The right answer for the question that is being asked and shown above is that: "B. 270° counterclockwise rotation around point G." the pre-image CDEF transformed to create image CꞌDꞌEꞌFꞌ is that B. 270° counterclockwise rotation around point G

The correct answer for this question is A. 180° clockwise rotation around point G!!

Hope i helped look at file below: