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What is the area of a parallelogram whose verticles are A(-12, 2), B (6, 2), C(-2, -3), and D (-20, -3)?

Liono4ka [1.6K]

The area of the parallelogram will be 108 square units.

<h3>What is the area of the parallelogram?</h3>

The space occupied by any two-dimensional figure in a plane is called the area. The area of the outer surface of any body is called the surface area.

The vertices of the parallelogram are (x₁, y₁), (x₂, y₂), (x₃, y₃), and ( x₄, y₄).

Then the area of the parallelogram will be calculated as below:-

Area = 1/2 |[(x₁y₂ + x₂y₃ + x₃y₄+x₄y₁) - (y₁x₂ + y₂x₃ + y₃x₄+y₄x₁)]|

We have

(x₁, y₁) ⇒ (-12, 2)

(x₂, y₂) ⇒ (6, 2)

(x₃, y₃) ⇒ (-2, -3)

(x₄, y₄) ⇒ (-20 , -3 )

Then the area will be

Area = 1/2[{(-12) x (2) + (6) x (-3) + (-2) x (-3) + (-20 x 2 )} – {(2) x (6) + (2) x (-2) + (-3) x (-20) + ( -3 ) ( -12 )}]

Area = 108 / 2

Area = 54 square units.

Therefore, the area of the parallelogram will be 108 square units.

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2 years ago

Geologists estimate the time since the most recent cooling of a mineral by counting the number of uranium fission tracks on the

Oliga [24]

Answer:

(a) The value of P (X = 7) is 0.1388.

(b) The value of P (X ≥ 3) is 0.9380.

(c) The value of P (2 < X < 7) is 0.5433.

(d) $\mu_{X}=6$

(e) $\sigma_{X}=2.45$

Step-by-step explanation:

Let <em>X</em> = number of uranium fission tracks on per cm² surface area of the mineral.

The average number of track per cm² surface area is, <em>λ</em> = 6.

The random variable <em>X</em> follows a Poisson distribution with parameter <em>λ</em> = 6.

The probability mass function of a Poisson distribution is:

$P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0, 1, 2, 3...$

(a)

Compute the value of P (X = 7) as follows:

$P(X=6)=\frac{e^{-6}(6)^{7}}{7!}=\frac{0.0025\times 279936}{5040}=0.1388$

Thus, the value of P (X = 7) is 0.1388.

(b)

Compute the value of P (X ≥ 3) as follows:

P (X ≥ 3) = 1 - P (X < 3)

= 1 - P (X = 0) - P (X = 1) - P (X = 2)

$=1-\frac{e^{-6}(6)^{0}}{0!}-\frac{e^{-6}(6)^{1}}{1!}-\frac{e^{-6}(6)^{2}}{2!}\\=1-0.00248-0.01487-0.04462\\=0.93803\\\approx0.9380$

Thus, the value of P (X ≥ 3) is 0.9380.

(c)

Compute the value of P (2 < X < 7) as follows:

P (2 < X < 7) = P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6)

$=\frac{e^{-6}(6)^{3}}{3!}+\frac{e^{-6}(6)^{4}}{4!}+\frac{e^{-6}(6)^{5}}{5!}+\frac{e^{-6}(6)^{6}}{6!}\\=0.08924+0.13385+0.16062+0.16062\\=0.54433\\\approx0.5443$