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Vesnalui [34]
3 years ago
15

1. Add or Subtract. Show your work

Mathematics
1 answer:
Sav [38]3 years ago
4 0

QUESTION 1a.

We want to add

( - 2 {x}^{2}  - 4x + 13) + (12 {x}^{2}  + 2x - 25)

We group similar terms to get

- 2 {x}^{2}  + 12 {x}^{2}- 4x + 2x   - 25 + 13

Simplify:

- 10{x}^{2} - 2x  -12

QUESTION 1b.

The given function is:

(7x^2+4x-26)-(-7x^2-3x+15)

Expand the parenthesis to obtain,

7x^2+4x-26 + 7x^2 + 3x - 15

Group similar terms,

7x^2+7 {x}^{2}  + 4x + 3x-26  - 15

Simplify:

14{x}^{2}  + 7x-41

QUESTION 2

The given function is

f(n)=n^2-6n-16

The zeros of this function is given by;

n^2-6n-16 = 0

Split the middle term.

n^2-8n + 2n-16 = 0

Factor;

n(n - 8) + 2(n - 8) = 0

(n - 8)(n + 2) = 0

n - 8 = 0 \: or \: n + 2 = 0

n  = 8 \: or \: n =  - 2

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5530 mm is 553 cm and 5.53 m.

5530 \: mm =  \frac{5530}{10}    = 553 \: cm \\ 553cm =  \frac{553}{100 }    = 5.53 \: m

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The correct answer is of option B.

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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).

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Solve and check thanks
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