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Bess [88]
3 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]3 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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\sqrt{\dfrac{1}{n}(\sum^6_{i=1}(x_i-\overline{x})^2)}\\\\=\sqrt{\dfrac{1}{6}(\sum^6_{i=1}(x_i-10)^2)}\\\\=\sqrt{\dfrac{1}{6}\times(49+16+1+1+25+36)}=4.61880215352\approx4.62

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The percentage of cotton in material used to manufacture men's shirts follows.
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Answer:

Median = 34.55

Quartiles:

Q1 = 33.1

Q2 = 34.55

Q3 = 35.6

Inter Quartile range = 2.5

Step-by-step explanation:

Step 1

We arrange the given the following data from lowest to the highest

32.0, 32.1, 32.1, 32.2, 32.4, 32.5, 32.5, 32.5, 32.8, 32.8, 32.8, 32.8, 32.9, 32.9, 33.0, 33.0, 33.2, 33.2, 33.2, 33.2, 33.5, 33.6, 33.8, 33.8, 33.9, 34.0, 34.0, 34.1, 34.1, 34.3, 34.4, 34.5, 34.6, 34.6, 34.6, 34.7, 34.7, 35.0, 35.0, 35.1, 35.1, 35.2, 35.4, 35.5, 35.5, 35.5, 35.5, 35.6, 35.6, 36.1, 36.3, 36.4, 36.4, 36.4, 36.5, 36.6, 36.7, 36.7, 37.0, 37.1, 37.4, 37.4, 37.6, 37.8

a. Construct a stem-and-leaf display for the data.

Stem and leaf Display

Stem | Leaf

32 | 0,1,1,2,4,5,5,5,8,8,8,8,9,9

33 | 0,0,2,2,2,2,5,6,8,8,9

34 | 0,0,1,1,3,4,5,6,6,6,7,7

35 | 0,0,1,1,2,4,5,5,5,5,6,6,

36 | 1,3,4,4,4,5,6,7,7,

37 | 0,1,4,4,6,8

b. Calculate the median and quartiles of these data.

Number of terms = 64

1) Median = 1/2(n + 1)th value

n = 64

= 1/2(64 + 1)th

= 1/2(65)th

= 32.5 th value

This means it is between the 32nd and 33rd value

32nd = 34.5

33rd = 34.6

= 34.5 + 34.6/2

= 69.1/2

= 34.55

2) First Quartile

1/4(n + 1)th value

n = 64

= 1/4(64 + 1)th

= 1/4(65)th

= 16.5th value

This means it is between the 16th and 17th value

16th value = 33.0

17tj value = 33.2

= 33.0 + 33.2/2

= 66.2/2

= 34.55

Q1 --> 33.1

3)Second Quartile = Median

1/2(n + 1)th value

n = 64

= 1/2(64 + 1)th

= 1/2(65)th

= 32.5 th value

This means it is between the 32nd and 33rd value

32nd = 34.5

33rd = 34.6

= 34.5 + 34.6/2

= 69.1/2

= 34.55

Q2 --> 34.55

Third Quartile

3/4(n + 1)th value

n = 64

= 3/4(64 + 1)th

= 3/4(65)th

= 48.75 th value

This means it is towards the 49th value

32nd = 34.5

Hence,

Q3 --> 35.6

Inter Quartile range

Q3 - Q1

= 35.6 - 33.1

= 2.5

8 0
3 years ago
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