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2 years ago

Statistic - Use the values in the contingency table to solve the equation given.

Mathematics

1 answer:

Lubov Fominskaja [6]2 years ago

5 0

Answer:

Step-by-step explanation:

$\begin{array}{c|c|c|c}&D&E&Sum\\---&---&---&---\\A&10&20&30\\B&15&5&20\\C&30&15&45\\---&---&---&---\\Sum&55&40&95\\\end {array}\\\\\\1) P(E)=\dfrac{40}{95} =\dfrac{8}{19} \\\\2) P(B \cup D)=P(B)+P(D)-P(B \cap D)=\dfrac{20}{95} +\dfrac{55}{95} -\dfrac{15}{95} =\dfrac{60}{95} =\dfrac{12}{19} \\\\3) P(A \cap E)=\dfrac{20}{95}=\dfrac{4}{19} \\\\4) P(B | E)=\dfrac{P(B \cap E )}{P(E)} =\dfrac{\dfrac{5}{95} }{\dfrac{40}{95} } =\dfrac{5}{40}= \dfrac{1}{8} \\\\5) P(D|E)=0$

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A triangle is graphed in the coordinate plane. The vertices of the triangle have coordinates (–3, 1), (1, 1), and (1, –2). What

vladimir2022 [97]

Answer:

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Step-by-step explanation:

Let

$A(-3,1),B(1,1),C(1,-2)$

we know that

The perimeter of triangle is equal to

$P=AB+BC+AC$

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

step 1

Find the distance AB

$A(-3,1),B(1,1)$

substitute in the formula

$AB=\sqrt{(1-1)^{2}+(1+3)^{2}}$

$AB=\sqrt{(0)^{2}+(4)^{2}}$

$AB=4\ units$

step 2

Find the distance BC

$B(1,1),C(1,-2)$

substitute in the formula

$BC=\sqrt{(-2-1)^{2}+(1-1)^{2}}$

$BC=\sqrt{(-3)^{2}+(0)^{2}}$

$BC=3\ units$

step 3

Find the distance AC

$A(-3,1),C(1,-2)$

substitute in the formula

$AC=\sqrt{(-2-1)^{2}+(1+3)^{2}}$

$AC=\sqrt{(-3)^{2}+(4)^{2}}$

$AC=5\ units$

step 4

Find the perimeter

$P=AB+BC+AC$

substitute the values

$P=4+3+5=12\ units$

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