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Naily [24]
3 years ago
6

Solve for 3a: a + 4 = 4a + 2 3a

Mathematics
1 answer:
expeople1 [14]3 years ago
7 0

Answer:

x = 2/3

2/3*3 = 2

Step-by-step explanation:

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3 years ago
Hi can someone help me understand how to work this problem out? I'll mark brainliest. Thanks:)
Ierofanga [76]

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0.73001

Step-by-step explanation:

1 month = 730.01 hours

730.01 * $0.001

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3 years ago
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almond37 [142]

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3 years ago
Read 2 more answers
A Martian couple has children until they have 2 males (sexes of children are independent). Compute the expected number of childr
Ne4ueva [31]

Answer:

a) 6

b) 4

c) 3

Step-by-step explanation:

Let p be the probability of having a female Martian, and of course, 1-p the probability of having a male Martian.

To compute the expected total number of trials before 2 males are born, imagine an experiment simulating the fact that 2 males are born is performed n times.

Let ak be the number of trials performed until 2 males are born in experiment k. That is,

a1= number of trials performed until 2 males are born in experiment 1

a2= number of trials performed until 2 males are born in experiment 2

and so on.

If a1 + a2 + … + an = N

we would expect Np females.  

Since the experiment was performed n times, there 2n males (recall that the experiment stops when 2 males are born).

So we would expect 2n = N(1-p), or

N/n = 2/(1-p)

But N/n is the average number of trials per experiment, that is, the expectation.

<em>We have then that the expected number of trials before 2 males are born is 2/(1-p) where p is the probability of having a female. </em>

a)

Here we have the probability of having a male is half as likely as females. So

1-p = p/2 hence p=2/3

The expected number of trials would be

2/(1-2/3) = 2/(1/3) =6

This means <em>the couple would have 6 children</em>: 4 females (the first 4 trials) and 2 males (the last 2 trials).

b)

Here the probability of having a female = probability of having a male = 1/2

The expected number of trials would be

2/(1/2) = 4

This means<em> the couple would have 4 children</em>: 2 females (the first 2 trials) and 2 males (the last 2 trials).

c)

Here, 1-p = 2p so p=1/3

The expected number of trials would be

2/(1-1/3) = 2/(2/3) = 6/2 =3

This means<em> the couple would have 3 children</em>: 1 female (the first trial) and 2 males (the last 2 trials).

5 0
4 years ago
The question is in the picture
disa [49]
B because the other options are used by Scientific notation
5 0
4 years ago
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