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3 years ago

In the diagram m angle ACB = 61 Find m angle BCD

Mathematics

2 answers:

Marina86 [1]3 years ago

7 0

we are given that

angle(ACF)=90

angle(ACB)=61

sum of all angles along any line is 180

so, we get

angle(ACF)+angle(ACD)=180

we can plug value

90+angle(ACD)=180

angle(ACD)=90

now, we can use formula

angle(ACD)=angle(ACB)+angle(BCD)

now, we can plug values

and we get

90=61+angle(BCD)

90-61=61-61+angle(BCD)

angle(BCD)=29................Answer

Basile [38]3 years ago

5 0

Answer:

Step-by-step explanation:

In the given diagram,

m(∠ACB) + m(∠ACf) +m(∠FCE) = 180°

[Angles formed at point on a straight line]

It is given that m(∠ACB) = 61°

m(∠ACF) = 90°

Therefore, 61° + 90° + m(∠FCE) = 180°

151 + m(∠FCE) = 180

m(∠FCE) = 180 - 151

= 29°

Since m(∠FCE) = (∠BCD)

[vertical opposite angles]

Therefore, m(∠BCD) = 29°

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2) X and Y are jointly continuous with joint pdf

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From what I gather from your latest comments, the PDF is given to be

$f_{X,Y}(x,y)=\begin{cases}cxy&\text{for }0\le x,y \le1\\0&\text{otherwise}\end{cases}$

and in particular, f(x, y) = cxy over the unit square [0, 1]², meaning for 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. (As opposed to the unbounded domain, x ≤ 0 *and* y ≤ 1.)

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$\displaystyle\int_0^1\int_0^1 cxy\,\mathrm dx\,\mathrm dy=c\left(\int_0^1x\,\mathrm dx\right)\left(\int_0^1y\,\mathrm dy\right)=\frac c{2^2}=1\implies \boxed{c=4}$

(b) Get the marginal density of X by integrating the joint density with respect to y :

$f_X(x)=\displaystyle\int_0^1 4xy\,\mathrm dy=(2xy^2)\bigg|_{y=0}^{y=1}=\begin{cases}2x&\text{for }0\le x\le 1\\0&\text{otherwise}\end{cases}$

$f_Y(y)=\displaystyle\int_0^14xy\,\mathrm dx=\begin{cases}2y&\text{for }0\le y\le1\\0&\text{otherwise}\end{cases}$

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