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Afina-wow [57]
3 years ago
13

Calculate the area of the triangle with the following vertices (3, -7), (6, 4), (-2, -3)

Mathematics
1 answer:
Monica [59]3 years ago
3 0

Answer:

\boxed{\mathsf{A} \triangle = \red{\dfrac{67}{2}u.a}}

Step-by-step explanation:

Let's follow up with the solution. Considering a triangle with the vertices \mathsf{A(x_A, y_A)}, \mathsf{B(x_B, y_B)} and \mathsf{C(x_C, y_C)}, have a look at the representation in the cartesian plan.

From this representation we can say that the area (A) of a triangle through the knowledge of <u>analytical geometry</u> is given by the determinant of the vertices divided by two, mathematically,

\mathsf{A} \triangle =  \dfrac{\left| \begin{array}{ccc}  \mathsf{x_A} & \mathsf{y_A }& 1 \\  \mathsf{x_B} &  \mathsf{ y_B} & 1 \\ \mathsf{ x_C} &  \mathsf{ y_C} & 1 \end{array} \right|}{2}

So, applying this knowledge we're going to have,

\mathsf{A} \triangle =  \dfrac{\left| \begin{array}{ccc}  3 & -7 & 1 \\ 6 &  4 & 1 \\ -2 &  -3 & 1 \end{array} \right|}{2}

\mathsf{A} \triangle =  \dfrac{1}{2}\left[  \left.\begin{array}{ccc}   3 & -7 & 1 \\ 6 &  4 & 1 \\ -2&  -3 & 1 \end{array}  \right| \begin{array}{cc} 3 & -7 \\ 6 & 4 \\ -2 & -3 \end{array} \right]

\mathsf{A} \triangle = \dfrac{12 + 14 - 18 - (-8 - 9 - 42)}{2}

\red{\mathsf{A} \triangle = \dfrac{67}{2} = 33,5u.a}

Hope you enjoy it, see ya!)

\green{\mathsf{FROM}}: Mozambique, Maputo – Matola City – T-3

DavidJunior17

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​41% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the
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Answer:

a) 0.2087 = 20.82% probability that the number of U.S. adults who have very little confidence in newspapers is exactly​ five.

b) 0.1834 = 18.34% probability that the number of U.S. adults who have very little confidence in newspapers is at least​ six.

c) 0.3575 = 35.75% probability that the number of U.S. adults who have very little confidence in newspapers is less than four.

Step-by-step explanation:

For each adult, there are only two possible outcomes. Either they have very little confidence in newspapers, or they do not. The answers of each adult are independent, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

​41% of U.S. adults have very little confidence in newspapers.

This means that p = 0.41

You randomly select 10 U.S. adults.

This means that n = 10

(a) exactly​ five

This is P(X = 5). So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 5) = C_{10,5}.(0.41)^{5}.(0.59)^{5} = 0.2087

0.2087 = 20.82% probability that the number of U.S. adults who have very little confidence in newspapers is exactly​ five.

(b) at least​ six

This is:

P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 6) = C_{10,6}.(0.41)^{6}.(0.59)^{4} = 0.1209

P(X = 7) = C_{10,7}.(0.41)^{7}.(0.59)^{3} = 0.0480

P(X = 8) = C_{10,8}.(0.41)^{8}.(0.59)^{2} = 0.0125

P(X = 9) = C_{10,9}.(0.41)^{9}.(0.59)^{1} = 0.0019

P(X = 10) = C_{10,10}.(0.41)^{10}.(0.59)^{0} = 0.0001

Then

P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.1209 + 0.0480 + 0.0125 + 0.0019 + 0.0001 = 0.1834

0.1834 = 18.34% probability that the number of U.S. adults who have very little confidence in newspapers is at least​ six.

(c) less than four.

This is:

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{10,0}.(0.41)^{0}.(0.59)^{10} = 0.0051

P(X = 1) = C_{10,1}.(0.41)^{1}.(0.59)^{9} = 0.0355

P(X = 2) = C_{10,2}.(0.41)^{2}.(0.59)^{8} = 0.1111

P(X = 3) = C_{10,3}.(0.41)^{3}.(0.59)^{7} = 0.2058

So

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0051 + 0.0355 + 0.1111 + 0.2058 = 0.3575

0.3575 = 35.75% probability that the number of U.S. adults who have very little confidence in newspapers is less than four.

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