What is the value of the expreson 4 tiny 4

Exercise 1:
exponential decay:
The function is given by:
y = A (b) ^ ((1/3) * t)
Where,
A = 600
We look for b:
(480/600) * (100) = 80%
b = 0.8
Substituting:
y = 600 * (0.8) ^ ((1/3) * t)
We check for t = 6
y = 600 * (0.8) ^ ((1/3) * 6)
y = 384
Answer:
exponential decay:
y = 600 * (0.8) ^ ((1/3) * t)
Exercise 2:
linear:
The function is given by:
y = ax + b
Where,
a = -60 / 2 = -30
b = 400
Substituting we have:
y = -30 * x + 400
We check for x = 4
y = -30 * 4 + 400
y = 280
Answer:
linear:
y = -30 * x + 400
Exercise 3:
exponential growth:
The function is given by:
y = A (b) ^ ((1/3) * t)
Where,
A = 512
We look for b:
(768/512) * (100) = 150%
b = 1.5
Substituting:
y = 512 * (1.5) ^ ((1/2) * t)
We check for t = 4
y = 512 * (1.5) ^ ((1/2) * 4)
y = 1152
Answer:
exponential growth:
y = 512 * (1.5) ^ ((1/2) * t)

8 0

3 years ago

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.

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.

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 the couple would have 6 children: 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 the couple would have 4 children: 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 the couple would have 3 children: 1 female (the first trial) and 2 males (the last 2 trials).

5 0

3 years ago