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Arada [10]
2 years ago
7

The answer appears in the image, this is for practice and please tell me an explanation on how you got the answer.

Mathematics
1 answer:
ale4655 [162]2 years ago
6 0
I hope this helps you :)

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A field is shaped like the figure shown. What is the area of the field? Use 3.14
uysha [10]
There is no figure, i cant help you sorry
6 0
3 years ago
Fine the slope of the line through each pair of points . (-8,-20),(5,2)
kozerog [31]

The formula of a slope:

m=\dfrac{y_2-y_1}{x_2-x_1}

We have the points (-8, -20) and (5, 2). Substitute:

m=\dfrac{2-(-20)}{5-(-8)}=\dfrac{22}{13}

Answer: The slope is \dfrac{22}{13}

7 0
3 years ago
Read 2 more answers
3. Using techniques from Calculus, show directly that the maximum value of a 1-D Gaussian distribution occurs at the point x = μ
Vilka [71]

Answer:

For a scaler variable, the Gaussian distribution has a probability density function of

p(x |µ, σ² ) = N(x; µ, σ² ) = 1 / 2π×e^{\frac{-(x-u)^{2}}{2s^{2} }  }

The term will have a maximum value at the top of the slope of the 1-D Gaussian distribution curve that is when exp(0) =1 or when x = µ

Step-by-step explanation:

Gaussian distributions have similar shape, with the mean controlling the location and the variance controls the dispersion  

From the graph of the probability distribution function it is seen that the the peak is the point at which the slope = 0, where µ = 0 and σ² = 1 then solution for the peak = exponential function = 0 or x = µ

5 0
3 years ago
How would I solve for x on this question
gregori [183]

Answer:

use your distributive properties and combine the two equations and just use pemdas.

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
The slope-intercept form of the equation of a line that passes through point (-3, 8) is y = -2/3x + 6. What is the point-
Elena-2011 [213]

y = \stackrel{\stackrel{m}{\downarrow }}{-\cfrac{2}{3}}x+6\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}

so hmmm then we know the slope of that line is -2/3, so we're really looking for the point-slope form of a line with a slope of -2/3 and that passes through (-3 , 8)

(\stackrel{x_1}{-3}~,~\stackrel{y_1}{8})\hspace{10em} \stackrel{slope}{m} ~=~ -\cfrac{2}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{8}=\stackrel{m}{-\cfrac{2}{3}}(x-\stackrel{x_1}{(-3)})\implies y-8=-\cfrac{2}{3}(x+3)

3 0
1 year ago
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