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vekshin1
3 years ago
13

What is the GCF of 2+18+40

Mathematics
2 answers:
tiny-mole [99]3 years ago
7 0
2 my friend..........
vlabodo [156]3 years ago
6 0
2 is the GCF of <span>2+18+40
.</span>
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What is the value of x?
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Given right triangle MNL what is the value of cos(M)?
FromTheMoon [43]

Step 1

<u>Find the length of the side MN</u>

we know that

Applying the Pythagorean Theorem

ML^{2} =NL^{2} +MN^{2}

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in this problem

ML=25\ units\\ NL=15\ units

Substitute in the formula above

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Step 2

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we know that

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cos (M)=\frac{adjacent\ side\ angle\ M}{hypotenuse}

adjacent\ side\ angle\ M=MN=20\ units\\ hypotenuse=ML=25\ units

Substitute

cos (M)=\frac{20}{25}

cos (M)=\frac{4}{5}

cos (M)=0.8

therefore

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The value of cos(M) is equal to 0.8


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3 years ago
You are traveling in a car. Your speed is 15 miles per hour. What is your speed in feet per minute?
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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
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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
3 years ago
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