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2 years ago

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16. For quadrilateral ABCD, determine the most precise name for it. A (-2, 3), B (9, 3), C (5, 6) and D (2, 6). Show your work a

nd explain.

2 answers:

ss7ja [257]2 years ago

6 0

Answer:

Trapzoid

Step-by-step explanation:

Given: A (-2, 3), B (9, 3), C (5, 6) and D (2, 6)

We will find the slope of each line.

Formula:

$\text{Slope, m}=\dfrac{y_2-y_1}{x_2-x_1}$

$\text{Slope of AB, m}_1=\dfrac{3-3}{9+2}=0$

$\text{Slope of BC, m}_2=\dfrac{6-3}{5-9}=-\dfrac{3}{4}$

$\text{Slope of CD, m}_3=\dfrac{6-6}{2-5}=0$

$\text{Slope of AD, m}_4=\dfrac{6-3}{2+2}=\dfrac{3}{4}$

Slope of AB = Slope of CD = 0

$m_1=m_3=0$

Thus, AB is parallel to CD

Slope of BC ≠ Slope of AD

$m_2\neq m_4$

Thus, BC is not parallel to AD

The quadrilateral ABCD has two sides are parallel and two are not parallel.

Hence, The quadrilateral is trapzoid

Genrish500 [490]2 years ago

4 0

The quadrilateral ABCD, with vertices <span>A(-2,3), B(9,3), C(5,6) and D(2, 6), is a trapezoid.

By definition, a trapezoid is a quadrilateral which has two parallel sides (These are called "bases"), but the other sides are not parallels. </span>

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For each women, there are only two possible outcomes. Either they visit their doctors for a physical examination at least once in two years, or they do not. The probability of a woman visiting their doctor at least once in this period is independent of any other women. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

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

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.

70% of adult women visit their doctors for a physical examination at least once in two years.

This means that $p = 0.7$

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This means that $n = 20$

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This is:

$P(X < 14) = 1 - P(X \geq 14)$

In which

$P(X \geq 14) = P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)$

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

$P(X = 14) = C_{20,14}.(0.7)^{14}.(0.3)^{6} = 0.1916$

$P(X = 15) = C_{20,15}.(0.7)^{15}.(0.3)^{5} = 0.1789$

$P(X = 16) = C_{20,16}.(0.7)^{16}.(0.3)^{4} = 0.1304$

$P(X = 17) = C_{20,14}.(0.7)^{17}.(0.3)^{3} = 0.0716$

$P(X = 18) = C_{20,18}.(0.7)^{18}.(0.3)^{2} = 0.0278$

$P(X = 19) = C_{20,19}.(0.7)^{19}.(0.3)^{1} = 0.0068$

$P(X = 20) = C_{20,20}.(0.7)^{20}.(0.3)^{0} = 0.0008$

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$P(X < 14) = 1 - P(X \geq 14) = 1 - 0.6079 = 0.3921$

0.3921 = 39.21% probability that fewer than 14 of them have had a physical examination in the past two years.

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From the values found in item (a).

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0.107 = 10.7% probability that at least 17 of them have had a physical examination in the past two years.

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