SHOW how you would use the distributive property to solve: 51 x 7

Instructions: Match each equations with correct type of event.

Mazyrski [523]

The question is incomplete. Below you will find the missing contents.

The correct match of events with order are,

P(A)P(B|A) - Dependent event
P(A)+P(B) - Mutually exclusive events
P(A and B)/P(A) - Conditional events
P(A) . P(B) - Independent Events
P(A)+P(B) -P(A and B) - not Mutually exclusive events.

When two events A and B are independent then,

P(A and B)=P(A).P(B)

when A and B are dependent events then,

P(A and B) = P(A) . P(B|A)

When two events A and B are mutually exclusive events then,

P(A and B)=0

So, P(A or B) = P(A) + P(B) - P(A and B) = P(A) + P(B)

P(A) + P(B) = P(A or B)

When events are not mutually exclusive then the general relation is,

P(A or B) = P(A) + P(B) - P(A and B)

If the probability of the event B conditioned by A is given by,

$\mathrm{P(B|A)=\frac{P(A~and~B)}{P(A)}}$

Hence the correct match are -

$\mathrm{P(A)P(B|A)\rightarrow}$ Dependent event
$\mathrm{P(A)+P(B)}\rightarrow$ Mutually exclusive events
$\mathrm{\frac{P(A ~and ~B)}{P(A)}\rightarrow}$ Conditional events
$\mathrm{P(A) . P(B)\rightarrow}$ Independent Events
$\mathrm{P(A)+P(B) -P(A ~and ~B)\rightarrow}$ not Mutually exclusive events.

Learn more about Probability of Events here -

brainly.com/question/79654680

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5 0

2 years ago

What is the area in square units of the rectangle

Vsevolod [243]

$\bf \textit{area of a rectangle}\\\\
A=length\cdot width\qquad \qquad A=\cfrac{3}{7}\cdot \cfrac{2}{9}\implies A=\cfrac{6}{63}\implies A=\cfrac{2}{21}$

8 0

4 years ago

An acute triangle has sides measuring 10 cm and 16 cm. The length of the third side is unknown.

saveliy_v [14]

Answer: Choice B

12.5 < x < 18.9

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Explanation:

We have a triangle with these side lengths:

a = 10
b = 16
c = x = unknown

Let's assume that b = 16 is the largest side of this triangle.

By the converse of the pythagorean theorem, we need $b^2 < a^2+c^2$ to be true in order for an acute triangle to happen.

So,

$b^2 < a^2 + c^2\\\\c^2 > b^2 - a^2\\\\c > \sqrt{b^2-a^2}\\\\x > \sqrt{16^2-10^2}\\\\x > \sqrt{156}\\\\x > 12.4899959967968 \ \text{(approximate)}\\\\x > 12.5$