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marta [7]
2 years ago
14

The back of Monique’s property is a creek. Monique would like to enclose a rectangular area, using the creek as one side and fen

cing for the other three sides, to create a corral. If there is 180 feet of fence available, what is the maximum possible area of the corral?
Mathematics
1 answer:
gogolik [260]2 years ago
3 0

Answer:

4050 sq. feet.

Step-by-step explanation:

Fencing is done on three sides of the rectangular area.

Given that there are 180 feet of fence available.

Then 2L + W = 180 ........(1), where L = length and W = width, of the rectangular plot.

Now, the area of the plot is given by A = LW

Now, from equation (1), we ger A = L (180 - 2L) ..... (2)

Then differentiating with respect to L in the both sides we get,

\frac{dA}{dL} = 180 - 4L =0 {Since condition for Area to be maximum is \frac{dA}{dL}=0}

⇒ L = 45 feet.

Now, from equation (2), we have A_{max} =L(180-2L) = 45(180 - 90) =4050 square feet.

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The Candela brothers own two pizza restaurants, one on Park Street and one on Bridge Road.
koban [17]

The mean, median and mode are measures of central tendency, that is they tend to indicate the location middle of the data

Required values;

(a) The performance for the week for Park Street

  • Revenue is <u>Q₂ < $7,500 < Q₃</u>
  • The sales for the week is better than <u>72.91%</u> of all sales

The performance for the week for Bridge Road

  • Revenue; <u>Q₂ < $7,100 < Q₃</u>
  • The sale for the week is better than <u>59.87%</u> of all sales

(b) The mean is <u>$3611</u>

The median is $<u>3,600</u>

The standard deviation is $<u>3250</u>

The Interquartile range is $<u>6075</u>

Reason:

The table of values that maybe used to find a solution to the question is given as follows;

\begin{array}{|l|l|l|}\mathbf{Variable} &\mathbf{Park}&\mathbf{Bridge}\\N&36&40\\Mean&6611&5989\\SE \ Mean&597&299\\StDev&3580&1794\\Minimum&800&1800\\Q_1&3600&5225\\Median&6600&6000\\Q_3&9675&7625\\Maximum&14100&8600\end{array}\right]

(a) Park Street revenue = $7,500

Bridge Road's revenue = $7,100

The two stores sold close to but below the 75th percentile

Bridge Road revenue;

The z-score is given as follows;

Z = \dfrac{x - \mu }{\sigma }

  • Z = \dfrac{7100 - 5,989 }{1794 } \approx 0.6193

From the Z-Table, we have;

The percentile= 0.7291

  • Therefore, the sale for the week for Park Street is better than <u>72.91%</u> of all the sales

Park Street revenue;

The z-score is given as follows;

  • Z = \dfrac{7500 - 6611}{3580} \approx 0.25

From the Z-Table, we have;

The percentile = <u>0.5987</u>

  • Therefore, the sale for the week is better than <u>59.87 %</u> of all the sales

(b) Given that the operating cost is $3,000, frim which we have;

The subtracted value is subtracted from the mean and median to find the new value

Profit = The revenue - Cost

New mean = 6611 - 3000 = 3611

  • The new mean = <u>$3,611</u>

The new median = 6600 - 3000 = 3600

  • The new median = <u>$3,600</u>

The standard deviation and the interquartile range remain the same, therefore, we have;

  • The standard deviation = <u>$3,580</u>

The interquartile range = 9675 - 3600 = 6075

  • The interquartile range = <u>6075</u>

Learn more here:

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Answer:

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Step-by-step explanation:

1. Let x and y represent the numbers of campers doing water- and ground-based activities, respectively.

  x + y ≤ 80 . . . . . . a maximum of 80 campers can be accommodated

  x ≥ 8 . . . . . . . . . . a minimum of 8 campers should be water-based

  y ≥ 5 . . . . . . . . . . a minimum of 5 campers should be ground-based

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2. See the attachment for a graph. Possible outcomes are shaded red. The boundary lines of the shaded area are included in the possible outcomes.

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3. Twenty campers are doing water-based activities and 40 campers are doing ground-based activities.

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Answer:

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Step-by-step explanation:

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