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sveta [45]
2 years ago
6

Problem PageQuestion An urn contains black and pink balls. Four balls are randomly drawn from the urn in succession, with replac

ement. That is, after each draw, the selected ball is returned to the urn. What is the probability that all balls drawn from the urn are pink? Round your answer to three decimal places.
Mathematics
1 answer:
Shkiper50 [21]2 years ago
8 0

Problem Page Question An urn contains black and pink balls. Four balls are randomly drawn from the urn in succession, with replacement. That is, after each draw, the selected ball is returned to the urn. What is the probability that all balls drawn from the urn are pink? Round your answer to three decimal places.

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A number divided by 6 minus -3 equals 12​
liberstina [14]

Let the number be x.

Therefore, by the problem

x/6-(-3) = 12

or, x/6+3 = 12

or, x/6 = 12-3

or, x/6 = 9

or, x = 9 × 6

or, x = 54

The number is 54.

4 0
2 years ago
Read 2 more answers
Nine times a number subtracted from another number​
ANTONII [103]

9x mean 9 multipled with an unkown number and 9x is subtracted from another unknown number 'y' . you can substitute with any letters you want ,but you have to use two different letters.

6 0
2 years ago
What is the Laplace Transform of 7t^3 using the definition (and not the shortcut method)
Leokris [45]

Answer:

Step-by-step explanation:

By definition of Laplace transform we have

L{f(t)} = L{{f(t)}}=\int_{0}^{\infty }e^{-st}f(t)dt\\\\Given\\f(t)=7t^{3}\\\\\therefore L[7t^{3}]=\int_{0}^{\infty }e^{-st}7t^{3}dt\\\\

Now to solve the integral on the right hand side we shall use Integration by parts Taking 7t^{3} as first function thus we have

\int_{0}^{\infty }e^{-st}7t^{3}dt=7\int_{0}^{\infty }e^{-st}t^{3}dt\\\\= [t^3\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(3t^2)\int e^{-st}dt]dt\\\\=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\

Again repeating the same procedure we get

=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt= \frac{3}{s}[t^2\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t^2)\int e^{-st}dt]dt\\\\=\frac{3}{s}[0-\int_{0}^{\infty }\frac{2t^{1}}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{2}}[\int_{0}^{\infty }te^{-st}dt]\\\\

Again repeating the same procedure we get

\frac{3\times 2}{s^2}[\int_{0}^{\infty }te^{-st}dt]= \frac{3\times 2}{s^{2}}[t\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t)\int e^{-st}dt]dt\\\\=\frac{3\times 2}{s^2}[0-\int_{0}^{\infty }\frac{1}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{3}}[\int_{0}^{\infty }e^{-st}dt]\\\\

Now solving this integral we have

\int_{0}^{\infty }e^{-st}dt=\frac{1}{-s}[\frac{1}{e^\infty }-\frac{1}{1}]\\\\\int_{0}^{\infty }e^{-st}dt=\frac{1}{s}

Thus we have

L[7t^{3}]=\frac{7\times 3\times 2}{s^4}

where s is any complex parameter

5 0
3 years ago
I need help solving this , help !
marin [14]

Answer:

NO

Step-by-step explanation:

Triangle A suggests that the triangle is isosceles.

Isosceles:

Triangle with two congruent angles and sides:

7 , 6 , 7

5 , 4 , 5

9, 12 , 9

Triangle B suggests that the triangle is scalene.

Scalene:

Triangle with no congruent angles or sides:

5 , 6 , 7

10 , 15 , 7

12 , 18 , 21

If my answer is incorrect, pls correct me!

If you like my answer and explanation, mark me as brainliest!

-Chetan K

6 0
3 years ago
If f(x) = 4x -11 what is the value of f(5)
sdas [7]

the answer is nine

hope this helps


7 0
3 years ago
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