Susan found the equation of the best fit line for the data shown in the scatterplot. The slope of the line of best fit is
1 answer:
Answer:
Negative
Step-by-step explanation:
From the scatter plot displayed, we could clearly observe the direction of the trend line as it produces a negative slope. For high values of y, the values of x are low and similarly, high values of x have low y values. Therefore, this kind of relationship between the two variables is considered negative.
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Answer:
The tax on property assessed at $660,000 is <u>$13860 </u>.
Step-by-step explanation:
Given:
In Atlanta the tax on a property assessed at $930,000 is $19,530.
The tax rates are proportional to this city.
Now, to find the tax on a property assessed at $660,000.
So, <em>we need to get the tax rate of Atlanta first.</em>
Tax Rate = Tax / Property assessed × 100.
<em>According to question the</em><u><em> tax rates are proportional.</em></u>
<em>So, </em><u><em>for getting the tax in this city we use the same rate of Atlanta</em></u><em>:</em>
Tax = tax rate of property assessed.
Therefore, the tax on property assessed at $660,000 is $13860.
Answer:
a. L{t} = 1/s² b. L{1} = 1/s
Step-by-step explanation:
Here is the complete question
The The Laplace Transform of a function ft), which is defined for all t2 0, is denoted by Lf(t)) and is defined by the improper integral Lf))s)J" e-st . f(C)dt, as long as it converges. Laplace Transform is very useful in physics and engineering for solving certain linear ordinary differential equations. (Hint: think of s as a fixed constant) 1. Find Lft) (hint: remember integration by parts) A. None of these. B. O C. D. 1 E. F. -s2 2. Find L(1) A. 1 B. None of these. C. 1 D.-s E. 0
Solution
a. L{t}
L{t} = ∫₀⁰⁰
Integrating by parts ∫udv/dt = uv - ∫vdu/dt where u = t and dv/dt = and v =
and du/dt = dt/dt = 1
So, ∫₀⁰⁰udv/dt = uv - ∫₀⁰⁰vdu/dt w
So, ∫₀⁰⁰ = [
]₀⁰⁰ - ∫₀⁰⁰
∫₀⁰⁰ = [
]₀⁰⁰ - ∫₀⁰⁰
= -1/s(∞exp(-∞s) - 0 × exp(-0s)) + [
]₀⁰⁰
= -1/s[(∞exp(-∞) - 0 × exp(0)] - 1/s²[exp(-∞s) - exp(-0s)]
= -1/s[(∞ × 0 - 0 × 1] - 1/s²[exp(-∞) - exp(-0)]
= -1/s[(0 - 0] - 1/s²[0 - 1]
= -1/s[(0] - 1/s²[- 1]
= 0 + 1/s²
= 1/s²
L{t} = 1/s²
b. L{1}
L{1} = ∫₀⁰⁰
= []₀⁰⁰
= -1/s[exp(-∞s) - exp(-0s)]
= -1/s[exp(-∞) - exp(-0)]
= -1/s[0 - 1]
= -1/s(-1)
= 1/s
L{1} = 1/s