Jacobi wants to install an underground sprinkler system in her backyard the backyard is rectangular with side length 17 m and 26 m .the water pipe will run diagonally across the yard about how many metres of water pipe does Jacobi need .

Answers

Answer 1

The length of the pipe required would be 31.06 meters

The length of the pipe is the hypotenus of the triangle formed :

hypotenus = √opposite² + adjacent²

substituting the values into our equation:

length of pipe = √17² + 26²

length of pipe = √965 = 31.06

Therefore, the length of the pipe needed is 31.06 meters

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Related Questions

A contour map is shown for a function f(x,y) on the rectangle R=[−3,6]×[−1,4]. a. Use the midpoint rule with m=2 and n=3 to estimate the value of ∬R​f(x,y)dA. b. Estimate the average value of the function f(x,y). fave​≈ Hint

Answers

a. The estimated value of ∬R​f(x,y)dA is 105

b. The estimated average value of the function f(x, y) is 7.

a. The rectangle R=[−3,6]×[−1,4] is divided into m = 2 subintervals along the x-axis and n = 3 subintervals along the y-axis. Therefore, each subinterval has a width of Δx = (6 - (-3))/2 = 9/2 and a height of Δy = (4 - (-1))/3 = 5/3.

We can calculate the midpoint of each subrectangle using the formula:

[tex]x_i = x_min + (i - 0.5) * \Delta x\\y_j = y_min + (j - 0.5) * \Delta y[/tex]

where i = 1, 2, ..., m and j = 1, 2, ..., n.

Using the midpoint rule, the estimate of the double integral is given by:

∬R​f(x,y)dA ≈ Δx * Δy * ∑∑[tex]f(x_i, y_j)[/tex]

where the double summation is taken over all the midpoints (x_i, y_j) of the subrectangles.

Calculate the midpoints of the subrectangles.

[tex]x_1 = -3 + (1 - 0.5) * (9/2) = -3 + 4.5 = 1.5\\x_2 = -3 + (2 - 0.5) * (9/2) = -3 + 9 = 6\\y_1 = -1 + (1 - 0.5) * (5/3) = -1 + (1/2) * (5/3) = -1 + 5/6 = -1/6\\y_2 = -1 + (2 - 0.5) * (5/3) = -1 + (3/2) * (5/3) = -1 + 5/2 = 9/2\\y_3 = -1 + (3 - 0.5) * (5/3) = -1 + (5/2) * (5/3) = -1 + 25/6 = 19/6[/tex]

Evaluate the function at each midpoint.

[tex]f(x_1, y_1) = 2\\f(x_1, y_2) = -1\\f(x_1, y_3) = 0\\f(x_2, y_1) = 1\\f(x_2, y_2) = 3\\f(x_2, y_3) = 2[/tex]

∬R​f(x,y)dA ≈ Δx * Δy * ∑∑[tex]f(x_i, y_j)[/tex]

           = (9/2) * (5/3) * (2 + (-1) + 0 + 1 + 3 + 2)

           = (9/2) * (5/3) * 7

           = 15 * 7

           =  105

b. To estimate the average value of the function f(x, y), we can divide the double integral by the area of the rectangle R, which is A = Δx * Δy * m * n.

The average value is then given by:

f_ave ≈ (∬R​f(x,y)dA) / A

Now let's perform the calculations:

Step 1: Calculate the area of the rectangle.

A = Δx * Δy * m * n

 = (9/2) * (5/3) * 2 * 3

 = 15

Step 2: Calculate the average value.

f_ave ≈ (∬R​f(x,y)dA) / A

     = 105 / 15

     = 7

Therefore, the estimated value of ∬R​f(x,y)dA is 105 and the estimated average value of the function f(x, y) is 7.

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The graph shows the function f(x) = |x – h| + k. What is the value of k?

Answers

The calculated value of k is -2.5

How to determine the value of k?

From the question, we have the following parameters that can be used in our computation:

The graph

(see attachment)

Also, we have

f(x) = |x - h| + k

From the graph, we have the vertex to be

(h, k) = (1, -2.5)

By comparison, we have

k = -2.5

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Let Y₁,..., Yn N(μ,0²). State the sampling distribution of Y = n=¹_₁ Y₁. -1 i=1 n1, Σ (Υ; – Υ)2. State the sampling distribution of S² = State the mean and variance of Y and S².

Answers

1. The sampling distribution of Y is a normal distribution with mean nμ and variance nσ².

2. The mean of the sampling distribution of S² is σ², and the variance is 2σ⁴ / (n-1).

In the given notation, Y₁, Y₂, ..., Yₙ are independent and identically distributed (i.i.d.) random variables following a normal distribution with mean μ and variance σ².

1. Sampling Distribution of Y = ∑(i=1 to n) Yᵢ:

The random variable Y represents the sum of n independent normal random variables. The sampling distribution of Y is also a normal distribution. The mean of the sampling distribution of Y can be obtained by the linearity of expectation:

E(Y) = E(∑(i=1 to n) Yᵢ) = ∑(i=1 to n) E(Yᵢ) = ∑(i=1 to n) μ = nμ

The variance of the sampling distribution of Y can be obtained by the linearity of variance:

Var(Y) = Var(∑(i=1 to n) Yᵢ) = ∑(i=1 to n) Var(Yᵢ) = ∑(i=1 to n) σ² = nσ²

Therefore, the sampling distribution of Y is a normal distribution with mean nμ and variance nσ².

2. Sampling Distribution of S²:

The random variable S² represents the sample variance calculated from a sample of n observations. The sampling distribution of S² follows a chi-square distribution with (n-1) degrees of freedom.The mean of the sampling distribution of S² is given by:

E(S²) = σ²

The variance of the sampling distribution of S² is given by:

Var(S²) = 2σ⁴ / (n-1)

Therefore, the mean of the sampling distribution of S² is σ², and the variance is 2σ⁴ / (n-1).

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