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RideAnS [48]
3 years ago
12

a clothing store office very often various promotions to attract customers but wants to maintain the same amount of profit to do

this the store mark up prices above retail and then advertises a deal that results in a markdown equivalent to a retail price the retail prices of a necktie is $59.99 the store offers a buy 2 get 1 free deal on neckties what is the new retail price during this promotion what is the markup and what is the markup percent on a tie during the promotion.
Mathematics
1 answer:
vova2212 [387]3 years ago
4 0

Since 3 neckties will be out from the store, so you need to know how much is the cost of 3 ties.

$59.99 x 3 = $179.97 and divide it by two because the costumer will only pay for 2

$179.97 / 2 = $89.99 is the new price

Mark up = 89.99 – 59.99 = $30.00

% mark up = [(89.99/59.99) -1] * 100

= 50%

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135% as a fraction or a mixed number fully simplified please
solong [7]

Answer:

135% = 27/20 = 17/20 as a fraction

Step-by-step explanation:

Step 1: Write down the percent divided by 100 like this:

135% = 135/100

Step 2: Multiply both top and bottom by 10 for every number after the decimal point:

As 135 is an integer, we don't have numbers after the decimal point. So, we just go to step 3.

Step 3: Simplify (or reduce) the above fraction by dividing both numerator and denominator by the GCD (Greatest Common Divisor) between them. In this case, GCD(135,100) = 5. So,

(135÷5)/(100÷5) = 27/20 when reduced to the simplest form.

As the numerator is greater than the denominator, we have an IMPROPER fraction, so we can also express it as a MIXED NUMBER, thus 135/100 is also equal to 1 7/20 when expressed as a mixed number.

6 0
2 years ago
Eight burgers and three orders of fries cost $79.49. Six orders of fries and twelve burgers cost $125.82.Find the price of one b
dezoksy [38]

Answer:

  • Burgers - $8.29 and fries - $4.39

Step-by-step explanation:

<u>Let burgers - b, fries - f</u>

  • 8b + 3f = 79.49
  • 12b + 6f = 125.82

<u>Double the first equation and subtract the second equation:</u>

  • 16b + 6f - 12b - 6f = 2(79.49) - 125.82
  • 4b = 33.16
  • b = 33.16/4
  • b = 8.29

<u>Find f:</u>

  • 8(8.29) + 3f = 79.49
  • 3f = 79.49 - 66.32
  • 3f = 13.17
  • f = 13.17/3
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5 0
2 years ago
Population Growth A lake is stocked with 500 fish, and their population increases according to the logistic curve where t is mea
Alexus [3.1K]

Answer:

a) Figure attached

b) For this case we just need to see what is the value of the function when x tnd to infinity. As we can see in our original function if x goes to infinity out function tend to 1000 and thats our limiting size.

c) p'(t) =\frac{19000 e^{-\frac{t}{5}}}{5 (1+19e^{-\frac{t}{5}})^2}

And if we find the derivate when t=1 we got this:

p'(t=1) =\frac{38000 e^{-\frac{1}{5}}}{(1+19e^{-\frac{1}{5}})^2}=113.506 \approx 114

And if we replace t=10 we got:

p'(t=10) =\frac{38000 e^{-\frac{10}{5}}}{(1+19e^{-\frac{10}{5}})^2}=403.204 \approx 404

d) 0 = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}

And then:

0 = 7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)

0 =19e^{-\frac{t}{5}} -1

ln(\frac{1}{19}) = -\frac{t}{5}

t = -5 ln (\frac{1}{19}) =14.722

Step-by-step explanation:

Assuming this complete problem: "A lake is stocked with 500 fish, and the population increases according to the logistic curve p(t) = 10000 / 1 + 19e^-t/5 where t is measured in months. (a) Use a graphing utility to graph the function. (b) What is the limiting size of the fish population? (c) At what rates is the fish population changing at the end of 1 month and at the end of 10 months? (d) After how many months is the population increasing most rapidly?"

Solution to the problem

We have the following function

P(t)=\frac{10000}{1 +19e^{-\frac{t}{5}}}

(a) Use a graphing utility to graph the function.

If we use desmos we got the figure attached.

(b) What is the limiting size of the fish population?

For this case we just need to see what is the value of the function when x tnd to infinity. As we can see in our original function if x goes to infinity out function tend to 1000 and thats our limiting size.

(c) At what rates is the fish population changing at the end of 1 month and at the end of 10 months?

For this case we need to calculate the derivate of the function. And we need to use the derivate of a quotient and we got this:

p'(t) = \frac{0 - 10000 *(-\frac{19}{5}) e^{-\frac{t}{5}}}{(1+e^{-\frac{t}{5}})^2}

And if we simplify we got this:

p'(t) =\frac{19000 e^{-\frac{t}{5}}}{5 (1+19e^{-\frac{t}{5}})^2}

And if we simplify we got:

p'(t) =\frac{38000 e^{-\frac{t}{5}}}{(1+19e^{-\frac{t}{5}})^2}

And if we find the derivate when t=1 we got this:

p'(t=1) =\frac{38000 e^{-\frac{1}{5}}}{(1+19e^{-\frac{1}{5}})^2}=113.506 \approx 114

And if we replace t=10 we got:

p'(t=10) =\frac{38000 e^{-\frac{10}{5}}}{(1+19e^{-\frac{10}{5}})^2}=403.204 \approx 404

(d) After how many months is the population increasing most rapidly?

For this case we need to find the second derivate, set equal to 0 and then solve for t. The second derivate is given by:

p''(t) = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}

And if we set equal to 0 we got:

0 = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}

And then:

0 = 7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)

0 =19e^{-\frac{t}{5}} -1

ln(\frac{1}{19}) = -\frac{t}{5}

t = -5 ln (\frac{1}{19}) =14.722

7 0
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What is the degree of the following polynomial? 4x^2+2x+7
adelina 88 [10]
The degree of the polynomial is 2.

to find the degree, put the polynomial in descending order with the highest exponents in front.  the highest exponent is the degree.

3x^6 - 4    (highest exponent here is 6, degree would be 6)

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BigorU [14]

Answer:

Step-by-step explanation:

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