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Marianna [84]
3 years ago
8

CAN SOMEONE PLEASE HELP FASTTT!!! 2 QUESTIONS FOR BRAINLIEST!!

Mathematics
1 answer:
11Alexandr11 [23.1K]3 years ago
7 0

Answer:

Question 1: seems all A,B and C are incorrect

Question 2: long = 45/(3*3) =45/9 =5

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Two non-simultaneous events have an equal likelihood of occurring (50%). If the events are independent of each other, what is th
BARSIC [14]
There are 4 possibilities:

1) No event happens
2) event A happens, not event B
3) event B happens, not event A
4) both events happen

seeing as there is a 50% chance of either event happening, I would assume that all 4 possibilities have an equal chance of happening

so if there are 4 events, then each one is 25% likely.

So the probability if them both occurring is 25%

I think
7 0
3 years ago
Read 2 more answers
Need now for study guide
likoan [24]

Answer:

17 cm.

Step-by-step explanation:

Because 12 cm + 5 cm. is 17  cm.

8 0
3 years ago
HELP ASAP PLEASE GIVING BRAINLIEST!!
Sholpan [36]

Answer:

$0.51 = 0.51 in decimal form

Step-by-step explanation:

3 0
3 years ago
5.15 divided by 0.24= I need an explanation
abruzzese [7]

Answer:

Step-by-step explanat

5.15=515/100

0.24=24/100

515/24=21.45

4 0
3 years ago
The prior probabilities for events A1 and A2 are P(A1) = 0.35 and P(A2) = 0.50. It is also known that P(A1 ∩ A2) = 0. Suppose P(
forsale [732]

Answer:

Step-by-step explanation:

Hello!

Given the probabilities:

P(A₁)= 0.35

P(A₂)= 0.50

P(A₁∩A₂)= 0

P(BIA₁)= 0.20

P(BIA₂)= 0.05

a)

Two events are mutually exclusive when the occurrence of one of them prevents the occurrence of the other in one repetition of the trial, this means that both events cannot occur at the same time and therefore they'll intersection is void (and its probability zero)

Considering that P(A₁∩A₂)= 0, we can assume that both events are mutually exclusive.

b)

Considering that P(BIA)= \frac{P(AnB)}{P(A)} you can clear the intersection from the formula P(AnB)= P(B/A)*P(A) and apply it for the given events:

P(A_1nB)= P(B/A_1) * P(A_1)= 0.20*0.35= 0.07

P(A_2nB)= P(B/A_2)*P(A_2)= 0.05*0.50= 0.025

c)

The probability of "B" is marginal, to calculate it you have to add all intersections where it occurs:

P(B)= (A₁∩B) + P(A₂∩B)=  0.07 + 0.025= 0.095

d)

The Bayes' theorem states that:

P(Ai/B)= \frac{P(B/Ai)*P(A)}{P(B)}

Then:

P(A_1/B)= \frac{P(B/A_1)*P(A_1)}{P(B)}= \frac{0.20*0.35}{0.095}= 0.737 = 0.74

P(A_2/B)= \frac{P(B/A_2)*P(A_2)}{P(B)} = \frac{0.05*0.50}{0.095} = 0.26

I hope it helps!

5 0
3 years ago
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