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Sonbull [250]
2 years ago
6

5. Find the selling Price of an item that originally cost a store $70 was marked up20% and then after no selling for 6 months wa

s discounted 10%?
Mathematics
1 answer:
zavuch27 [327]2 years ago
4 0

Answer:

$75.6

Explanation:

If the original cost was $70 and it was marked up 20%, the price after this is calculated as:

$70 + $70(0.2) = $70 + $14 = $84

Where $70(0.2) = $14 is the amount added to the original when it is marked up 20%.

Then, it was discounted 10%, we get that the selling price is

$84 - $84(0.1) = $84 - $8.4 = $75.6

Where $84(0.1) = $8.4 is the discount made on the initial price.

Therefore, the answer is $75.6

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Rama09 [41]
The associative property of addition is pretty much just any numbers in place of these: a + (b + c) = (a + b) + c. Therefore, your answer would be d) (6 + 1) + 9 = 6 + (1 + 9)
6 0
3 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
3 years ago
2.) Two students are working on a simplification problem. Their work is shown below. Is either of
kvasek [131]

Answer:

Both Jeff and Lori are wrong.

Step-by-step explanation:

3 -  \frac{(4 \times 3)^{2} }{2}  + (1 \times 6)

3 - \frac{12^{2}}{2}  + 6

3 -  \frac{144}{2}  + 6

3 - 72 + 6

- 72 + 9

- 63

-63 is the final answer

8 0
3 years ago
Please help me thank you I’ll mark Brainly
Inessa05 [86]

Answer:

the answer would be d

Step-by-step explanation:

(8+12) is the sum

(8+12)10 would give you the answer of the sum of 8 and 12 times 10

7 0
3 years ago
Read 2 more answers
To earn money for a contest, the math team pays $40 for a box of spirit banners. They sell each banner for $3.50. Write an equat
ankoles [38]

Answer:

3.5b - 40 where b is the amount of banners sold.

Step-by-step explanation:

We know the team has already spent $40, and they earn #3.5 per banner.

Hence, with b representing the amount of banners sold, the equation is 3.5b - 40

7 0
3 years ago
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