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Vanyuwa [196]
3 years ago
11

Any suggestions or ideas

Mathematics
2 answers:
forsale [732]3 years ago
8 0
From one point to the other there is a horiz. distance of 5, and a vert dist. of 12.

then D = sqrt(5^2 + 12^2) = 13 units (answer)
kirill115 [55]3 years ago
4 0
That will be D good sir
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Arnoldo needs to write this system in slope-intercept form. Which shows how he could do that? 3 x minus 2 y = 6. 0.4 (20 y + 15)
kotegsom [21]

Answer:

  y = three-halves x minus 3

Step-by-step explanation:

Subtract 3x from both sides of the original equation.

  -2y = -3x +6

Divide by -2

  y = 3/2x -3

4 0
3 years ago
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1 yard =3 feet and 1 foot =12 inches
AysviL [449]

Answer:

54 yds=162 ft, 1944 in

Step-by-step explanation:

Ⓗⓘ ⓣⓗⓔⓡⓔ

Well, 54 yds*3 ft=162 ft

162 ft*12 in=1944 in

(っ◔◡◔)っ ♥ Hope this helped! Have a great day! :) ♥

Please, please give brainliest, it would be greatly appreciated, I only a few more before I advance, thanks!

3 0
3 years ago
FIRST ANSWER GETS BRAINLIEST (ANSWER HAS TO BE IN DETAIL)
agasfer [191]

Answer:

2975.978

Step-by-step explanation:

This is pretty easy, just keep the order going on.

Here is the CORRECT order:

2 Thousand(s)

9 Hundred(s)

7 Ten(s)

5 One(s)

decimal part because there is <em>th</em> after each type of number (ex. thousand<em>th)</em>

9 Tenth(s)

7 Hundredth(s)

8 Thousandth(s)

3 0
2 years ago
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Determine the x- and y-intercepts of the graph of y=−1/3x+3 .
andrey2020 [161]
X-intercept : -9 (-9,0)
y-intercept: 3 (0,3)
3 0
3 years ago
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Find the cdf F(x) associated with each of the following probability density functions. Sketch the graphs of f(x) and F(x).
wolverine [178]

Answer:

See steps below

Step-by-step explanation:

a)

\bf f(x)=3(1-x)^2\;(0

\bf F(x)=\int_{0}^{x}f(t)dt=\int_{0}^{x}3(1-t)^2dt=3\int_{0}^{x}(1-t)^2=1-(1-x)^3

The cdf associated with f is

\bf \boxed{F(x)=1-(1-x)^3} for 0<x<1

<h3>See picture 1 </h3>

The median is a point x such that

F(x) = ½

so, the median is

\bf 1-(1-x)^3=1/2\rightarrow (1-x)^3=1/2\rightarrow \boxed{x=1-\sqrt[3]{2}}

The 25th percentile equals the 1st quartile and is a point x such

F(x) = ¼

and the 25th percentile is

\bf 1-(1-x)^3=1/4\rightarrow (1-x)^3=3/4\rightarrow \boxed{x=1-\sqrt[3]{3/4}}

b)

\bf f(x)=\frac{1}{x^2}\;(1

\bf F(x)=\int_{1}^{x}f(t)dt=\int_{1}^{x}\frac{dt}{t^2}=1-\frac{1}{x}

The cdf associated with f is

\bf \boxed{F(x)=1-\frac{1}{x}} for x>1

<h3>See picture 2 </h3>

The median is

\bf 1-\frac{1}{x}=\frac{1}{2}\rightarrow \boxed{x=2}

The 25th percentile is  

\bf 1-\frac{1}{x}=\frac{1}{4}\rightarrow \boxed{x=4/3}

c)

f(x) = 1/3 for 0<x<1 or 2<x<4

\bf F(x)=\int_{0}^{x}\frac{dt}{3}=\frac{x}{3}\;(0

\bf F(x)=\frac{1}{3}+\int_{2}^{x}\frac{dt}{3}=\frac{1}{3}+\frac{x-2}{3}=\frac{x-1}{3}\;(2

The cdf associated with f is

\bf F(x)=\frac{x}{3} for 0<x<1

\bf F(x)=\frac{x-1}{3} for 2<x<4

<h3>See picture 3 </h3>

The median is

\bf \frac{x-1}{3}=1/2\rightarrow x=1+3/2\rightarrow \boxed{x=5/2}

The 25th percentile is  

\bf \frac{x}{3}=1/4\rightarrow \boxed{x=3/4}

4 0
4 years ago
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