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svet-max [94.6K]
4 years ago
10

In ΔUVW, w = 44 cm, u = 83 cm and ∠V=141°. Find the area of ΔUVW, to the nearest square centimeter.

Mathematics
1 answer:
Tema [17]4 years ago
8 0

Answer:

Area  \approx 1149\ cm^2

Step-by-step explanation:

<u>Given that:</u>

ΔUVW,

Side w = 44 cm, (It is the side opposite to \angle W)

Side u = 83 cm (It is the side opposite to \angle U)

and ∠V=141°

Please refer to the attached image with labeling of the triangle with the dimensions given.

Area of a triangle with two sides given and angle between the two sides can be formulated as:

A = \dfrac{1}{2}\times a\times b\times sinC

Where a and b are the two sides and

\angle C is the angle between the sides a and b

Here we have a = w = 44cm

b = u = 44cm

and ∠C= ∠V=141

Putting the values to find the area:

A = \dfrac{1}{2}\times 44\times 83\times sin141\\\Rightarrow A = \dfrac{1}{2} \times 3652 \times sin141\\\Rightarrow A =1826 \times 0.629\\\Rightarrow A  \approx 1149\ cm^2

So, the <em>area </em>of given triangle to the nearest square centimetre is:

Area  \approx 1149\ cm^2

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E(X) = \sum_{i=1}^n X_i P(X_i) = 3*0.07 +4*0.4 +5*0.25 +6*0.28= 4.74In order to find the variance we need to calculate first the second moment given by:

E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 3^2*0.07 +4^2*0.4 +5^2*0.25 +6^2*0.28= 23.36And the variance is given by:

Var(X) = E(X^2) +[E(X)]^2 = 23.36 -[4.74]^2 = 0.8924

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Step-by-step explanation:

Previous concepts

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).  

Solution to the problem

For this case we have the following distribution given:

X          3      4       5        6

P(X)   0.07  0.4  0.25  0.28

We can calculate the mean with the following formula:

E(X) = \sum_{i=1}^n X_i P(X_i) = 3*0.07 +4*0.4 +5*0.25 +6*0.28= 4.74

In order to find the variance we need to calculate first the second moment given by:

E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 3^2*0.07 +4^2*0.4 +5^2*0.25 +6^2*0.28= 23.36

And the variance is given by:

Var(X) = E(X^2) +[E(X)]^2 = 23.36 -[4.74]^2 = 0.8924

And the deviation would be:

Sd(X) =\sqrt{0.8924} =0.9447

3 0
4 years ago
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