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Mariana [72]
3 years ago
9

This programmer can type 55 words per minute At that rate how many words can the programmer type in 11 minutes A.66. B.550. C.60

5. D.655
Mathematics
2 answers:
AveGali [126]3 years ago
3 0
Take 1 min = 55 words as a conversion factor. In order to achieve the words that a programmer could type, multiply the given 11 minutes to the conversion factor. Thus, it is viewed as (55 words/ 1 minute ) x 11 minutes in order to cancel some units, the answer would be 605 words.
Lady_Fox [76]3 years ago
3 0
Umm you go to k12
cause i go there

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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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4 years ago
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Eduardwww [97]

Answer:

The function f(x) has a vertical asymptote at x = 3

Step-by-step explanation:

We can define an asymptote as an infinite aproximation to given value, such that the value is never actually reached.

For example, in the case of the natural logarithm, it is not defined for x = 0.

Then Ln(x) has an asymptote at x = 0 that tends to negative infinity, (but never reaches it, as again, Ln(x) is not defined for x = 0)

So a vertical asymptote will be a vertical tendency at a given x-value.

In the graph is quite easy to see it, it occurs at x = 3 (the graph goes down infinitely, never actually reaching the value x = 3)

Then:

The function f(x) has a vertical asymptote at x = 3

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Answer:

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Step-by-step explanation:

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