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3 years ago

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Mathematics

1 answer:

Monica [59]3 years ago

4 0

Answer:

a. Length = 13 yards and Width = 21 yards.

b. Area of rectangle = 273 square yards.

Step-by-step explanation:

Let the length of the rectangular garden be L

Let the width of the rectangular garden be W

Given the following data;

Perimeter of garden = 68 yards

Translating the word problem into an algebraic equation, we have;

W = 2L - 5 ......equation 1

Note: The formula for calculating the perimeter of a rectangle is;

$P = 2L + 2W$

68 = 2L + 2W ........equation 2

Substituting eqn 1 into eqn 2;

68 = 2L + 2(2L - 5)

68 = 2L + 4L - 10

68 = 6L - 10

6L = 68 + 10

6L = 78

L = 78/6

L = 13 yards

To find the width;

W = 2L - 5

W = 2(13) - 5

W = 26 - 5

W = 21 yards.

b. To find the area of the garden;

Area of rectangle = length * width

Area of rectangle = 13 * 21

Area of rectangle = 273 square yards.

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Gala2k [10]

Answer:

(a) There are 70 different ways set up 4 computers out of 8.

(b) The probability that exactly three of the selected computers are desktops is 0.305.

Step-by-step explanation:

Of the 9 new computers 4 are laptops and 5 are desktop.

Let X = a laptop is selected and Y = a desktop is selected.

The probability of selecting a laptop is = $P(Laptop) = p_{X} = \frac{4}{9}$

The probability of selecting a desktop is = $P(Desktop) = p_{Y} = \frac{5}{9}$

Then both X and Y follows Binomial distribution.

$X\sim Bin(9, \frac{4}{9})\\ Y\sim Bin(9, \frac{5}{9})$

The probability function of a binomial distribution is:

$P(U=k)={n\choose k}\times(p)^{k}\times (1-p)^{n-k}$

(a)

Combination is used to determine the number of ways to select k objects from n distinct objects without replacement.

It is denotes as: ${n\choose k}=\frac{n!}{k!(n-k)!}$

In this case 4 computers are to selected of 8 to be set up. Since there cannot be replacement, i.e. we cannot set up one computer twice or thrice, use combinations to determine the number of ways to set up 4 computers of 8.

The number of ways to set up 4 computers of 8 is:

${8\choose 4}=\frac{8!}{4!(8-4)!}\\=\frac{8!}{4!\times 4!} \\=70$

Thus, there are 70 different ways set up 4 computers out of 8.

(b)

It is provided that 4 computers are randomly selected.

Compute the probability that exactly 3 of the 4 computers selected are desktops as follows:

$P(Y=3)={4\choose 3}\times(\frac{5}{9})^{3}\times (1-\frac{5}{9})^{4-3}\\=4\times\frac{125}{729}\times\frac{4}{9}\\ =0.304832\\\approx0.305$

Thus, the probability that exactly three of the selected computers are desktops is 0.305.

(c)

Compute the probability that of the 4 computers selected at least 3 are desktops as follows:

$P(Y\geq 3)=1-P(Y$

Thus, the probability that at least three of the selected computers are desktops is 0.401.

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2 years ago

Which is the equation for ---

ratelena [41]

Equation of the line:
y=mx+b
Ordinate at the origin (where the line cut the y-axis): b=1

B=(-5,5)=(xb,yb)→xb=-5, yb=5
C=(0,1)=(xc,yc)→xc=0, yc=1
m=(yc-yb)/(xc-xb)
m=(1-5)/(0-(-5))
m=(-4)/(0+5)
m=(-4)/(5)
m=-(4/5)

y=mx+b
y=-(4/5)x+1
Multiplying the equation by 5:
5[y=-(4/5)x+1]
5y=-4x+5
Adding 4x both sides of the equation:
5y+4x=-4x+5+4x
4x+5y=5

Answer: The equation for BC is: Third option 4x+5y=5

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2 years ago

What is the answer to this question 2x + x + x + 2y x 3y x y

nikklg [1K]

Answer:

2 x (3 y^3 + 2)

Step-by-step explanation:

Simplify the following:

2 x + x + x + 2×3 y y x y

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2 x + x + x + 2×3 y^2 x y

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2 x + x + x + 6 y^2 x y

6 y^2 x y = 6 y^(2 + 1) x:

2 x + x + x + 6 y^(2 + 1) x

2 + 1 = 3:

2 x + x + x + 6 y^3 x

Grouping like terms, 2 x + x + x + 6 y^3 x = 6 x y^3 + (2 x + x + x):

6 x y^3 + (2 x + x + x)

2 x + x + x = 4 x:

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3 years ago

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Kryger [21]

15/2 if you are binging it to simplest form

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Butoxors [25]

Answer:

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Step-by-step explanation:

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2 years ago