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GrogVix [38]
3 years ago
6

Consider an experiment in which the letters "A"

Mathematics
1 answer:
spin [16.1K]3 years ago
6 0

0.2 iis the answer you are loooking for when you say pal is not formed

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Maksim231197 [3]

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4 0
3 years ago
Read 2 more answers
How to solve this question
Yuki888 [10]
If the radical, is the same in this case ✔️10, simply add or subtract the whole numbers or the coefficients of the radicals regularly, the solution would be 1✔️10 or simply ✔️10.
4 0
3 years ago
Holy has 12 feet of ribbon. if she uses 3.5 feet to wrap one gift ,what is the maximum length of ribbon holly can use to wrap th
skelet666 [1.2K]
Hey!

OK, so we need to find out how much ribbon she will have left in the end to use for the rest of her gifts.

12 - 3.5 = 8.5

The maximum length of ribbon Holly can use to wrap the rest of her gifts is 8.5 feet. 

Hope this helps! 
6 0
4 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).

a.

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
3 years ago
Adult tickets to space amusement parkcost x dollars. Children's tickets cost y dollars. The henson family bought 3 aduly and 1 c
vesna_86 [32]

Answer:

3x + 1y = $163

2x + 3y =$174

Step-by-step explanation:

Adult tickets to space amusement park cost x dollars children’s tickets cost y dollars.

Henson family bought 3 adults and 1 child's tickets for $163

x + y = 163

3x + 1y = 163

The Garcia family bought 2 adults and 3 child’s ticket for $174.

x + y = $174

2x + 3y = $174

Therefore the equations representing both families cost are

Henson family

3x + 1y = $163

Garcia family

2x + 3y = $174

3 0
4 years ago
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