What is the slope of the line shown?
2 answers:
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The walkway is 1.5 m wide.
The area of the pool is 12(6) = 72 m².
Adding a walkway of unknown width, x, around all 4 sides of the pool increases the width by 2x and the length by 2x; thus the area of the entire pool and walkway together would be given by
(12+2x)(6+2x)
We know that the area of just the walkway is 9 m² less than the area of the pool. This means that:
(12+2x)(6+2x)-72 = 72-9
Multiplying through we have:
12*6+12*2x+2x*6+2x*2x - 72 = 63
72 + 24x + 12x + 4x² - 72 = 63
24x + 12x + 4x² = 63
36x + 4x² = 63
Writing in standard form we have:
4x² + 36x = 63
We want to set it equal to 0 to solve, so subtract 63 from both sides:
4x² + 36x - 63 = 63 - 63
4x² + 36x - 63 = 0
Using the quadratic formula,
Since a negative width makes no sense, the walkway is 1.5 m wide.
The area of the pool is 12(6) = 72 m².
Adding a walkway of unknown width, x, around all 4 sides of the pool increases the width by 2x and the length by 2x; thus the area of the entire pool and walkway together would be given by
(12+2x)(6+2x)
We know that the area of just the walkway is 9 m² less than the area of the pool. This means that:
(12+2x)(6+2x)-72 = 72-9
Multiplying through we have:
12*6+12*2x+2x*6+2x*2x - 72 = 63
72 + 24x + 12x + 4x² - 72 = 63
24x + 12x + 4x² = 63
36x + 4x² = 63
Writing in standard form we have:
4x² + 36x = 63
We want to set it equal to 0 to solve, so subtract 63 from both sides:
4x² + 36x - 63 = 63 - 63
4x² + 36x - 63 = 0
Using the quadratic formula,
Since a negative width makes no sense, the walkway is 1.5 m wide.
Answer:
The new mean is 5.
The new standard deviation is also 2.
Step-by-step explanation:
Let the sample space of hours be as follows, S = {x₁, x₂, x₃...xₙ}
The mean of this sample is 4. That is,
The standard deviation of this sample is 2. That is, .
Now it is stated that each of the sample values was increased by 1 hour.
The new sample is: S = {x₁ + 1, x₂ + 1, x₃ + 1...xₙ + 1}
Compute the mean of this sample as follows:
The new mean is 5.
Compute the standard deviation of this sample as follows:
The new standard deviation is also 2.