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4 years ago

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Event A is a subset of event B. P(A)=.15 and P(B)=.53 Find P(A|B). Round the answer to the nearest hundredth.

1 answer:

ludmilkaskok [199]4 years ago

4 0

If $A\subseteq B$ , then $A\cap B=A$ . So

$\mathbb P(A\mid B)=\dfrac{\mathbb P(A\cap B)}{\mathbb P(B)}=\dfrac{\mathbb P(A)}{\mathbb P(B)}=\dfrac{0.15}{0.53}\approx0.28$

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Answer:

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3 years ago

Read 2 more answers

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4 years ago

The coordinates of ∆ABC are A(-3, 2), B(5, 8) & C(11, 0). Which type of triangle is ∆ABC? Select All that apply.

stepan [7]

Given:

The coordinates of ∆ABC are A(-3, 2), B(5, 8) & C(11, 0).

To find:

The type of the given triangle.

Solution:

Distance formula:

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using the distance formula, we get

$AB=\sqrt{(5-(-3))^2+(8-2)^2}$

$AB=\sqrt{(8)^2+(6)^2}$

$AB=\sqrt{64+36}$

$AB=\sqrt{100}$

$AB=10$

Similarly,

$BC=\sqrt{(11-5)^2+(0-8)^2}$

$BC=\sqrt{(6)^2+(8)^2}$

$BC=\sqrt{36+64}$

$BC=\sqrt{100}$

$BC=10$

And,

$AC=\sqrt{(11-(-3))^2+(0-2)^2}$

$AC=\sqrt{(14)^2+(-2)^2}$

$AC=\sqrt{196+4}$

$AC=\sqrt{200}$

$AC=10\sqrt{2}$

Two sides of the triangle are equal, i.e., $AB=BC$ . So, the triangle is an isosceles triangle.

Sum of square of two smaller side is

$AB^2+BC^2=10^2+10^2$

$AB^2+BC^2=100+100$

$AB^2+BC^2=200$

$AB^2+BC^2=AC^2$

Using the Pythagoras theorem, we can say that the given triangle is a right triangle.

Therefore, the correct options are B and F.

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3 years ago