The management of the unico department store has decided to enclose an 834 ft2 area outside the building for displaying potted plants and flowers. one side will be formed by the external wall of the store, two sides will be constructed of pine boards, and the fourth side will be made of galvanized steel fencing. if the pine board fencing costs $6/running foot and the steel fencing costs $3/running foot, determine the dimensions of the enclosure that can be erected at minimum cost. (round your answers to one decimal place.)

Answers

Answer 1
Answer: The area is given by:
 A = x * y = 834
 The cost equation is given by:
 C = 6 * (2x) + 3 * (y)
 We express the equation in terms of a variable:
 C (x) = 6 * (2x) + 3 * (834 / x)
 Rewriting:
 C (x) = 12x + 2502 / x
 We derive the equation:
 C '(x) = 12 - 2502 / x ^ 2
 We match zero:
 0 = 12 - 2502 / x ^ 2
 We clear x:
 x = root ((2502) / (12))
 x = 14.4 feet
 We look for the other dimension:
 y = 834 / x
 y = 834 / 14.4
 y = 57.9 feet
 Answer:
 
The dimensions of the enclosure that can be erected at minimum cost are:
 
x = 14.4 feet
 
y = 57.9 feet

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Decrease 520 by 15% what Is the answer?

Answers

The number is 442 which is obtained after decreasing 520 by 15% the answer is 442.

What is the percentage?

It's the ratio of two integers stated as a fraction of a hundred parts. It is a metric for comparing two sets of data, and it is expressed as a percentage using the percent symbol.

Let x be the number that is obtained after decreasing 520 by 15%

= (100 - 15)% of 520

= 85% of 520

= 0.85×520

= 442

Thus, the number is 442 which is obtained after decreasing 520 by 15% the answer is 442.

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Answer:

442

Step-by-step explanation:

If you want to decrease 520 by 15%, you have to multiply 15% by 520 first, which is 78. Next, you have to minus 78 from 520. Now you get 442!

(a) Count the number of ways to select a sample of 3 people to serve on a board of directors from a population of 6 people. answer: (b) If a simple random sampling procedure is to be employed, the chances that any particular sample will be the one selected are

Answers

Answer:

a) 20

b) Equally likely, 0.05

Step-by-step explanation:

We are given the following in the question:

Population size, n = 6

Sample size, r = 3

a) Ways to select a sample of 3 from a population of 6

= n^C_r\n=^6C_3\n\n=(6!)/(3!(6-3)!)\n\n=(6!)/(3!3!)\n\n=20

Thus, there are 20 ways in which a sample of 3 can be selected from a population of 6.

b) The chances that any particular sample will be the one selected are equally likely.

\text{Probability} = \displaystyle\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}

Probability of selecting one particular sample  =

\text{P(Sample)} = (1)/(20) = 0.05

Thus, 0.05 is the equally likely probability of selecting one sample.

. A coin is tossed three times, and the sequence of heads and tails is recorded.(a) Determine the sample space, Ω.(b) List the elements that make up the following events: i.A= exactly two tails, ii.B= at least twotails, iii.C= the last two tosses are heads(c) List the elements of the following events: i.A, ii.A∪B, iii.A∩B, iv.A∩C

Answers

Answer:

See explanation below

Step-by-step explanation:

Here a coin was tossed three times.

Let H = head &  T = tail

Find the following:

a) The sample space:

Since a coin is tossed thrice, all possible outcome would be:

S = { HHH, HHT, HTH, HTT, TTT, TTH, THH, THT}

b) i) A = Exactly 2 tails: Here exactly 2 tails were recorded.

A = {HTT, TTH, THT}

ii) B = at least two tails: Here 2 or more tails were recorded.

B = {HTT, TTT, TTH, THT}

iii) C = the last two tosses are heads:

C = { HHH, THH}

c) List the elements of the following events:

i) A. This means all outcomes in A

= {HTT, TTH, THT}

ii) A∪B. A union B, means all possible outcomes present in A or B or in both

= {HTT, TTH, THT, TTT}

iii) A∩B. This means all possible outcomes of A that are present in B.

= {HTT, TTH, THT}

iv) A∩C. All outcomes A that are present in B

= {∅}

The sample space of tossing a coin three times consists of eight possible outcomes: HHH, HHT, HTH, THH, TTH, THT, HTT, and TTT. Events A, B, and C can be determined by listing the appropriate outcomes. The intersection and union of events A and B can also be determined.

(a) The sample space, Ω, of tossing a coin three times can be determined by listing all the possible outcomes: HHH, HHT, HTH, THH, TTH, THT, HTT, and TTT.

(b) i. A = {HHT, HTH, THH}

ii. B = {TTT, TTH, THT, HTT, HHT, HTH, THH}

iii. C = {HTH, TTH}

(c) i. A = {HHT, HTH, THH}

ii. A∪B = {HHT, HTH, THH, TTT, TTH, THT, HTT, HHT}

iii. A∩B = {HHT, HTH, THH}

iv. A∩C = {HHT, HTH}

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Twenty air-conditioning units have been brought in for service. Twelve of them have broken compressors, and eight have broken fans. Seven units are chosen at random to be worked on. What is the probability that three of them have broken fans

Answers

Answer: 0.3576

Step-by-step explanation:

Let the probability that the units that have broken fans from the seven units chosen at random be x.

The probability that three of them have broken fans will be 0.3576. Check the attachment for details

What is the radius of a sphere with a volume of 572 cm", to the nearest tenth of a
centimeter?

Answers

Answer:

The radius of a sphere with a volume of 572 cm^3 is 5.15 cm

Step-by-step explanation:

We need to find the radius of a sphere whose volume is: 572 cm^3

The formula used will be: Volume=(4)/(3) \pi r^3

Putting value of Volume =572 and π=3.14 we can find radius (r) of a sphere.

Volume=(4)/(3) \pi r^3\n572= (4)/(3)*3.14*r^3\n572=4.187*r^3\nr^3=(572)/(4.187)\nr^3=136.613\nTaking \ cube \ root \ on \ both \ sides:\n\sqrt[3]{r} =\sqrt[3]{136.613} \nr=5.15 \ cm

So, the radius of a sphere with a volume of 572 cm^3 is 5.15 cm

Answer:

its 5.1

Step-by-step explanation:

Body temperature​ (in degrees​ Fahrenheit) of randomly selected normal and healthy adults are shown below. Compute the​ mean, median, and mode of the data set.98.1
98.6
98.7
98.5
98.0
98.2
98.0
99.0
98.0
98.8

The mean is °F.

​(Round to the nearest hundredth as​ needed.)

Answers

The mean is the average median is the middle value of the data set while the mode is the highest frequency number thus mean median and mode are 98.39,98.35 and 98 respectively.

What are the mode and median?

Mode is the highest frequency number while the median is the middle value of a data set after writing in either an increasing or decreasing manner.

The increasing order of the given dataset,

98.0,98.0,98.0,98.1,98.2,98.5,98.6,98.7,98.8,99.0

Mean

(98.0+98.0+98.0+98.1+98.2+98.5+98.6+98.7+98.8+99.0 )/10

⇒ 98.39

Median

(98.2+98.5)/2 = 98.35

Mode

The highest frequency of 98.

Hence "The mean is the average median is the middle value of the data set while the mode is the highest frequency number thus mean median and mode are 98.39,98.35 and 98 respectively".

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Answer:

Mean: 98.49

Median: 98.35

Mode: 98