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3 years ago

Find the missing length in the right triangle 6.5 and 9.9

Mathematics

1 answer:

Oduvanchick [21]3 years ago

3 0

Answer:

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Joe made a scale drawing of the community pool in his town. the pool has a perimeter of 77 meters, what is the length and width

viva [34]

Answer:

The length of the pool is 28 meters and the width of the pool is 10.5 meters.

Step-by-step explanation:

In the scale drawing of Joe, the community pool has (length : width) = 8 : 3

Let the actual length of the pool is 8x meters and the actual width is 3x meters.

Now, given that the actual pool has a perimeter of 77 meters.

So, 2(8x + 3x) = 77

⇒ 22x = 77

⇒ x = 3.5

So, the length of the pool is 8x = 8(3.5) = 28 meters and the width of the pool is 3x = 3(3.5) = 10.5 meters. (Answer)

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3 years ago

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4 years ago

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You are purchasing a car for $12,465.00 plus 5.65% sales tax. You make a $1,300.00 down payment and have a fair credit score. If

pochemuha

Answer:

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Step-by-step explanation:

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3 years ago

The picture shows 3 triangles. Triangle 2 and Triangle 3 are images of Triangle 1 under

ch4aika [34]

The side lengths are the same.
Step-by-step explanation:
A rigid transformation does not change any dimensions. The side lengths are the same.

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3 years ago

Given a joint PDF, f subscript X Y end subscript (x comma y )equals c x y comma space 0 less than y less than x less than 4, (1)

ioda

(1) Looks like the joint density is

$f_{X,Y}(x,y)=\begin{cases}cxy&\text{for }0$

In order for this to be a proper density function, integrating it over its support should evaluate to 1. The support is a triangle with vertices at (0, 0), (4, 0), and (4, 4) (see attached shaded region), so the integral is

$\displaystyle\int_0^4\int_y^4 cxy\,\mathrm dx\,\mathrm dy=\int_0^4\frac{cy}2(4^2-y^2)=32c=1$

$\implies\boxed{c=\dfrac1{32}}$

(2) The region in which X > 2 and Y < 1 corresponds to a 2x1 rectangle (see second attached shaded region), so the desired probability is

$P(X>2,Y$

(3) Are you supposed to find the marginal density of X, or the conditional density of X given Y?

In the first case, you simply integrate the joint density with respect to y:

$f_X(x)=\displaystyle\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dy=\int_0^x\frac{xy}{32}\,\mathrm dy=\begin{cases}\frac{x^3}{64}&\text{for }0$

In the second case, we instead first find the marginal density of Y:

$f_Y(y)=\displaystyle\int_y^4\frac{xy}{32}\,\mathrm dx=\begin{cases}\frac{16y-y^3}{64}&\text{for }0$

Then use the marginal density to compute the conditional density of X given Y:

$f_{X\mid Y}(x\mid y)=\dfrac{f_{X,Y}(x,y)}{f_Y(y)}=\begin{cases}\frac{2xy}{16y-y^3}&\text{for }y$

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3 years ago

Find the missing length in the right triangle 6.5 and 9.9​

Find the missing length in the right triangle 6.5 and 9.9