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3 years ago

A = {a, b, c, d}

Mathematics

1 answer:

Helen [10]3 years ago

4 0

Answer:

Johnny is wrong.

$A \cup B = \left\{a,b,c,d,e,f,g,h\right\}$

Step-by-step explanation:

Johnny is wrong.

A better definition would be: $A \cup B$ is the set of all elements that belong to at least one A or B.

So, the elements that belong to both A and B, like c and d in this exercise, also belong to $A \cup B$ .

So:

$A \cup B=\left\{a,b,c,d,e,f,g,h\right\}$

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For the animal shelter, Elena Nppurchased

Naddik [55]

Elena's total cost (C) is 5(9.98 + 24.99 + x)

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Total cost (C) = 5 * $9.98 + (5 * 24.99) + (5 * x)

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Therefore Elena's total cost (C) is 5(9.98 + 24.99 + x)

Find out more at: brainly.com/question/21105092

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2 years ago

If A and B are independent events with P(A) = .5 and P(B) = .2, find the following:a) P(A U B)b) P(A^c ? B^c)c) P(A^c U B^c)**No

Effectus [21]

If A and B are independent, then $P(A\cap B)=P(A)P(B)$ .

$P(A\cup B)=P(A)+P(B)-P(A\cap B)=P(A)+P(B)-P(A)P(B)$

$P(A\cup B)=0.5+0.2-0.5\cdot0.2$

$\boxed{P(A\cup B)=0.6}$

b. I'm guessing the ? is supposed to stand for intersection. We can use DeMorgan's law for complements here:

$P(A^c\cap B^c)=P(A\cup B)^c=1-P(A\cup B)$

$P(A^c\cap B^c)=1-0.6$

$\boxed{P(A^c\cap B^c)=0.4}$

c. DeMorgan's law can be used here too:

$P(A^c\cup B^c)=P(A\cap B)^c=1-P(A\cap B)=1-P(A)P(B)$

$P(A^c\cup B^c)=1-0.5\cdot0.2$

$\boxed{P(A^c\cup B^c)=0.9}$

4 0

2 years ago

UESTION 6

Xelga [282]

Explique melhor, se eu entender consigo te ajuda :)

4 0

3 years ago

Item 8

SCORPION-xisa [38]

53 minutes

3:08 to 4:00 is 52 minutes.

to 4:01 would add an extra minute which would be 53.

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3 years ago

Read 2 more answers

What kind of number is 0.87 <br><br> rational or irrational

grigory [225]

Answer: it is rational

Step-by-step explanation:

7 0

2 years ago