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Bess [88]
4 years ago
8

An owner of a key rings manufacturing company found that the profit earned (in thousands of dollars) per day by selling n number

of key rings is given by , where n is the number of key rings in thousands. Find the number of key rings sold on a particular day when the total profit is $5,000.
n^2-2n-3
Mathematics
1 answer:
vovangra [49]4 years ago
3 0

Answer:

The number of key rings sold on that day is 4000 key rings

Step-by-step explanation:

* Lets explain the information in the problem

- The profit earned (in thousands of dollars) per day by selling n number

  of key rings is given by the function P(n) = n² - 2n - 3, where n is the

  number of key rings in thousands and P is the profit in thousands

  for one day

- On a particular day the total profit is $5,000

∵ 5000 = 5 in thousands

∵ The function P(n) is the profit of n key ring in thousands

∴ P(n) = 5

- Lets solve the function to find the number of key rings

∵ P(n) = n² - 2n - 3

∴ 5 = n² - 2n - 3 ⇒ subtract 5 from both sides

∴ 0 = n² - 2n - 8 ⇒ factorize it

∵ n² = n × n ⇒ 1st terms in the 2 brackets

∵ -8 = -4 × 2 ⇒ 2nd terms in the 2 brackets

∵ n × -4 = -4n ⇒ nears

∵ n × 2 = 2n ⇒ extremes

∵ -4n + 2n = -2n ⇒ the middle term

∴ (n - 4)(n + 2) = 0 ⇒ equate each bracket by 0 to find n

∴ n - 4 = 0 ⇒ add 4 to both sides

∴ n = 4 key ring in thousands = 4000 key rings

- OR

∴ n + 2 = 0 ⇒ subtract 2 from both sides

∴ n = -2 ⇒ we will refused this value because number of key rings

   must be positive

∴ The number of key rings sold on that day is 4000 key rings

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Mark brainliest please


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6 0
3 years ago
What is the greatest common factor of 16ab3 + 4a2b + 8ab ?
ziro4ka [17]

Answer:

2ab(3b^2+2a+4)

Step-by-step explanation:

6ab^3 + 4a^2b + 8ab

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Factor out the common terms

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5 0
3 years ago
A probability experiment is conducted in which the sample space of the experiment is upper s equals startset 4 comma 5 comma 6 c
olchik [2.2K]
The correct question statement is:

A probability experiment is conducted in which the sample space of the experiment is S = {4,5,6,7,8,9,10,11,12,13,14,15}. Let event E={7,8,9,10,11,12,13,14,15}. Assume each outcome is equally likely. List the outcomes in E^{c}. Find P(E^{c}).

Solution:

Part 1:

E^{c} means compliment of the set E. A compliment of a set can be obtained by finding the difference of the set from the universal set. The universal set is the set which contains all the possible outcomes of the events which is S in this case.

So, compliment of E will be equal to S - E. S - E will result in all those elements of S which are not present in E. So, we can write:

E^{c}=S-E \\  \\ 
E^{c}=(4,5,6,7,8,9,10,11,12,13,14,15)-(7,8,9,10,11,12,13,14,15) \\  \\ 
E^{c}=(4,5,6)

Thus the set compliment of E will contain the elements {4,5,6}.So

E^{c} = {4,5,6}

Part 2)

P(E^{c}) means probability that if we select any number from the Sample Space S, it will belong the set E compliment.

P(E^{c}) = (Number of Elements in E^{c})/Number of elements in S

Number of elements in set S = n(S) = 12
Number of elements in set E^{c} = n(E^{c})=3

So, 

P(E^{c})= \frac{n(E^{c}) }{n(S)} \\  \\ 
P(E^{c})= \frac{3}{12} \\  \\ 
P(E^{c})= \frac{1}{4}
7 0
3 years ago
When y = 7, what is the value of x?
GREYUIT [131]

Answer:-1 as said in the bottom of the equation

Step-by-step explanation: it is at the bottom of the equation if wrong sorry.

3 0
3 years ago
The greater rnge is the boys or the grls or neither as it is the same
maks197457 [2]
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median of boys = 12.5
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Hope it helps!

5 0
3 years ago
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