2 years ago

Find the value of x. Then classify the triangle by its angles.

Mathematics

2 answers:

Pani-rosa [81]2 years ago

6 0

X= 15° and it is and acute triangle

nikklg [1K]2 years ago

3 0

Answer:

it is an acute triangle

Step-by-step explanation:

explanation is it's all under 90 degrees angles

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(C) -4a +7b
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aivan3 [116]

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Tomtit [17]

Given:

The frequency distribution table.

To find:

The mean average score on a test.

Solution:

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$Marks (x_i)$ $Frequency(f_i)$ $f_ix_i$

x a xa

y b yb

z c zc

Sum a+b+c xa+yb+zc

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$Mean=\dfrac{\sum f_ix_i}{\sum f_i}$

$Mean=\dfrac{xa+yb+zc}{a+b+c}$

Therefore, the mean average score on the test is $\dfrac{xa+yb+zc}{a+b+c}$ .

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2 years ago

Suppose u1, u2, ..., un are independent random variables and for every i = 1, ..., n, ui has a uniform distribution over [0, 1].

sattari [20]

$Z=U_{(1)}=\min\{U_1,\ldots,U_n\}$

has CDF

$F_Z(z)=1-(1-F_{U_i}(z))^n$

where $F_{U_i}(u_i)$ is the CDF of $U_i$ . Since $U_i$ are iid. with the standard uniform distribution, we have

$F_{U_i}(u_i)=\begin{cases}0&\text{for }u_i$

and so

$F_Z(z)=1-(1-F_{U_i}(z))^n=\begin{cases}0&\text{for }z$

Differentiate the CDF with respect to $z$ to obtain the PDF:

$f_Z(z)=\dfrac{\mathrm dF_Z(z)}{\mathrm dz}=\begin{cases}n(1-z)^{n-1}&\text{for }0$

i.e. $Z$ has a Beta distribution $\beta(1,n)$ .

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2 years ago