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Tanzania [10]
2 years ago
12

A rectangular field is to be enclosed by a fence and divided

Mathematics
1 answer:
Phantasy [73]2 years ago
5 0

Answer:

80000 square meters

Step-by-step explanation:

perimeter + dividing fence = 800

let a = length

let b = width

let c = length of dividing fence

perimeter = 2*a + 2*b

let's say...

c is the same as the length

2a + 2b + c = 800

2a + 2b + a = 800

3a + 2b = 800

area = length*width

area = a*b

area / b = a

3*(area/b) + 2b = 800

3*(area/b) = 800 - 2b

area/b = (800 - 2b)

area = (800 - 2b)*b

To make the area large, we make the right hand side large.

800b - 2b^2

If you put in terms of x, y it looks like a downward opening parabola, so the max area is at the vertex. Half way between the roots.

y = -2x^2 + 800x

y = -x^2 + 400x

0 = x*(-x + 400)

roots are x= 0 and x = 400

vertex is at x, aka b = 200

area at b=200 is (800 - 400)*200 = 80000

and a is area/b... 80000/400 = 200

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Answer:

the second option and x=20

4 0
3 years ago
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Please solve the whole page please please
Leya [2.2K]

1d) ↑ (Do the same but turned if that's easier)

1a) 199

1b) 193

1c) 217

1e) Based on the box plot, we can for example, find the lowest and greatest values, the range (largest - smallest), the mean (all values / # values), and the median middle value or q2 given by the second line in the box.

2b) 7

2c) 9.5

2d) ↑

2e) because the data is not ordered, it is difficult to determine the quartile data. Once it is ordered like 6, 7, 7, 7, 8, 8, 9, 10, 10. The data can be visualized alot easier. With the box plot we can determine q3 of the data by looking at the 3rd line, using the whiskers of the box to find the minimum and maximum values.

3)

You need to find the minimum, maximum, the quartile data obtained by having an even amount of data on both sides, order it so that values can be grouped. this includes q1, q2 (or median), and q3.

In this case the ordered data would be: 4,6,8,8,9,11,12,14,14,16.

[4, 6, 8, 8, 9] [] [11, 12, 14, 14, 16]

4 is the minimum, 16 is the maximum, q1 is 8, q3 is 14, and median is 10.

4) ↑

5)

You need to find the minimum, maximum, the quartile data obtained by having an even amount of data on both sides, order it so that values can be grouped. this includes q1, q2 (or median), and q3.

In this case the ordered data would be:

45, 45, 50, 55, 60, 70, 70, 75, 80, 85

[45, 45, 50, 55, 60] [] [70, 70, 75, 80, 85]

45 is the minimum, 85 is the maximum, q1 is 50, q3 is 75, and median is 65.

6) ↑

4 0
2 years ago
Plz i need help with this. It would really help if you help me with this (._.)
Ira Lisetskai [31]
Z is B and x is D. Hope this helps!








6 0
3 years ago
Three cards are drawn at random from a standard deck of cards, without replacement. Find the probability that the suits of the c
Evgesh-ka [11]

The probability that suits of cards are spades, hearts, and spades is 39/850 or 0.046.

According to the give question.

Three cards are drawn from a standard deck of cards without replacement.

Since,

The total number of cards in a standard deck = 52

Total number of spade cards = 13

Total number of heart cards = 13

Now,

The probabbility of getting first card as a spade card = 13/52 = 1/4

The probability of getting second card as heart card

= 13/(52 -1)     (cards are drawn without replacement)

= 13/51

The probability of getting third card as a spade card

= (13 - 1)/52 -2

= 12/50

= 6/25

Therefore, the probability that suits of the cards are spades, hearts, spades

= 1/4 × 13/51 × 6/25

= 39/850

= 0.046

Hence, the probability that suits of cards are spades, hearts, and spades is 39/850 or 0.046.

Find out more information about probability here:

brainly.com/question/11234923

#SPJ4

8 0
1 year ago
EEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE EEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE
mestny [16]

Answer:

This man is valid

5 0
3 years ago
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