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Dmitry_Shevchenko [17]
3 years ago
12

Suppose i pick a jelly bean at random from a box containing two red and ten blue ones. i record the color and put the jelly bean

back in the box. if i do this three times, what is the probability of getting a blue jelly bean each time? (round your answer to three decimal places.)
Mathematics
1 answer:
gtnhenbr [62]3 years ago
8 0
The probability of picking a blue jelly bean is 10/12=0.833, since there are 10 blue and 12 jelly beans in total.

Each time the probability of picking blue is the same, since put back in the box whatever jelly bean we pick

P(blue, blue, blue) = P(blue) × P(blue) × P(blue) = 0.833×0.833×0.833=0.579


Answer: 0.579
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Answer:

A=34\ units^2

Step-by-step explanation:

Suppose we have a general triangle like the one shown in the figure.

We know the angle A, the angle B and the length b.

A = 30\°\\\\B = 45\°\\\\b = 10

By definition I know that the sum of the internal angles of a triangle is always equal to 180 °.

So

A + B + C = 180\\\\30 + 45 + C = 180

We solve the equation and thus we find the angle C.

C = 180 - 30-45\\\\C = 105

We already know the three triangle angles.

Now we use the sine theorem to calculate the sides c and a.

The  sine theorem says that:

\frac{sin(A)}{a}=\frac{sin(B)}{b}=\frac{sin(C)}{c}

Then

\frac{sin(30)}{a}=\frac{sin(45)}{10}

\frac{sin(30)}{\frac{sin(45)}{10}}=a

a=7.071

Also

\frac{sin(105)}{c}=\frac{sin(45)}{10}

\frac{sin(105)}{\frac{sin(45)}{10}}=c

c=13.660

Finally, we use the Heron formula to calculate the triangle area

A=\sqrt{s(s-a)(s-b)(s-c)}

Where s is:

s=\frac{a+b+c}{2}

Therefore

s=\frac{7.071+10+13.660}{2}

s=15.37

A=\sqrt{15.37(15.37-7.071)(15.37-10)(15.37-13.66)}

A=34\ units^2

3 0
3 years ago
Given that f(x) = 2x + 5 and g(x) = x − 7, solve for f(g(x)) when x = −3.
WITCHER [35]
G(-3) = -3-7 = -10;
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Step-by-step explanation:

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(x² + 5) (x − 2) (x −3)

The zeros are 2, 3, and ±√5i.

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3 years ago
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Step-by-step explanation:

Here we are given that , the Value of

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And we need to find the prove that ,

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We know a identity as ,

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On cubing both sides of the given equation,

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Simplify and then substitute ,

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Hence Proved !

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3 years ago
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