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3 years ago

Solve for x leave answer as decimal 6/5=x/3

Mathematics

2 answers:

777dan777 [17]3 years ago

8 0

Answer:

3 3/5 =x

3 .6 =x

Step-by-step explanation:

6/5 = x/3

Use cross products

6*3 = 5*x

18 = 5x

Divide each side by 5

18/5 = 5x/5

18/5 =x

3 3/5 =x

Dmitrij [34]3 years ago

8 0

Answer: 3.6 = x

Step-by-step explanation: To solve for x in the given proportion, we simply use the means-extremes property which states the the product of the extremes, (6)(3), equals the product of the means, (5)(x).

So we have (6)(3) or 18 equals (5)(x) or 5x.

So we have the equation 18 = 5x.

Now dividing both sides by 5, <em>3.6 = x</em>.

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Find the probability that a randomly generated bit string of length 10 does not contain a 0 if bits are independent and if a 0 b

jeka94

Answer:

Step-by-step explanation:

From the given information; Let's assume that R should represent the set of all possible outcomes generated from a bit string of length 10 .

So; as each place is fitted with either 0 or 1

$\mathbf{|R|= 2^{10}}$

Similarly; the event E signifies the randomly generated bit string of length 10 does not contain a 0

Now;

if a 0 bit and a 1 bit are equally likely

The probability that a randomly generated bit string of length 10 does not contain a 0 if bits are independent and if a 0 bit and a 1 bit are equally likely is;

$\mathbf{P(E) = \dfrac{|E|}{|R|}}$

so ; if bits string should not contain a 0 and all other places should be occupied by 1; Then:

$\mathbf{{|E|}=1 }$ ; $\mathbf{|R|= 2^{10}}$

$\mathbf{P(E) = \dfrac{1}{2^{10}}}$

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2 years ago

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Alecsey [184]

Answer:

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Step-by-step explanation:

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2 years ago

Read 2 more answers

Is x+y+1=0 a tangent of both y^2=4x and x^2=4y parabolas?

Lubov Fominskaja [6]

Answer:

yes

Step-by-step explanation:

The line intersects each parabola in one point, so is tangent to both.

For the first parabola, the point of intersection is ...

y^2 = 4(-y-1)

y^2 +4y +4 = 0

(y+2)^2 = 0

y = -2 . . . . . . . . one solution only

x = -(-2)-1 = 1

The point of intersection is (1, -2).

For the second parabola, the equation is the same, but with x and y interchanged:

x^2 = 4(-x-1)

(x +2)^2 = 0

x = -2, y = 1 . . . . . one point of intersection only

___

If the line is not parallel to the axis of symmetry, it is tangent if there is only one point of intersection. Here the line x+y+1=0 is tangent to both y^2=4x and x^2=4y.

_____

Another way to consider this is to look at the two parabolas as mirror images of each other across the line y=x. The given line is perpendicular to that line of reflection, so if it is tangent to one parabola, it is tangent to both.