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Art [367]
3 years ago
12

There are 15 tennis balls in a box, of which 9 have not previously been used. three of the balls are randomly chosen, played wit

h, and then returned to the box. later, another 3 balls are randomly chosen from the box. find the probability that none of these balls has ever been used
Mathematics
2 answers:
vagabundo [1.1K]3 years ago
8 0
For the first three balls that are chosen, the probability would be found by the following expression: 9/15*8/14*7/13 = 12/65  The numerator decreases since the number of new balls is decreasing with each one that is removed. The denominator decreases since the total number of tennis balls in the box is being decreased. For the second time that balls are chosen, the following expression is used: 6/15*5/14*4/13 = 10/273 The numerator starts with 6 since there are now fewer balls that are unused due to the ones previously have been used. These two fractions are then multiplied to find the total probability that none of the 6 balls were used. 12/65*10/273 = 8/1183
siniylev [52]3 years ago
3 0
Total tennis balls = 15 balls
Previously not used balls = 9 balls
Previously used balls = 15 - 9 = 6 balls

Probability of getting first 3 balls that are not used previously = \frac{9}{15} ×  \frac{8}{14} ×  \frac{7}{13}

Now remaining total balls = 15 - 3 = 12
Balls that are not previously used = 9 - 3 = 6

Probability of getting next 3 balls that are not used previously = \frac{6}{12} ×  \frac{5}{11} ×  \frac{4}{10}

Probability = \frac{9}{15} ×  \frac{8}{14} ×  \frac{7}{13} × \frac{6}{12} ×  \frac{5}{11} ×  \frac{4}{10}

= 0.089

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Elena-2011 [213]

To start to figure this problem out, we can first figure out how many more pages you read in an hour compared to your sister.


50 - 40 = 10, so you read 10 more pages in 1 hour.


Now we have to figure out how much more you would read in 1/2 an hour. Since 1/2 an hour is 1/2 of an hour, we can divide 10 by 2 to figure out how many more pages you read in an hour compared to your sister. 10 divided by 2 is 5, so you read 5 more pages every half hour.

6 0
3 years ago
Help please! ill mark brainliest
lesya [120]

Answer:

90deg

Step-by-step explanation:

If the angles are supplementary, that means mA + mB = 180*

we know that

mA = (2x + 8)deg

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mA + mB = 180deg

mB + mB = 2mB = 180deg

mB = 90deg

Note, that we did not need to compute x even, because the angles were the same.

6 0
3 years ago
What number is 9 times as much as 400 ? A 391 B 409 C 3,600 D 3,609​
Eddi Din [679]

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8 0
2 years ago
Lagrange multipliers have a definite meaning in load balancing for electric network problems. Consider the generators that can o
Ivahew [28]

Answer:

The load balance (x_1,x_2,x_3)=(545.5,272.7,181.8) Mw minimizes the total cost

Step-by-step explanation:

<u>Optimizing With Lagrange Multipliers</u>

When a multivariable function f is to be maximized or minimized, the Lagrange multipliers method is a pretty common and easy tool to apply when the restrictions are in the form of equalities.

Consider three generators that can output xi megawatts, with i ranging from 1 to 3. The set of unknown variables is x1, x2, x3.

The cost of each generator is given by the formula

\displaystyle C_i=3x_i+\frac{i}{40}x_i^2

It means the cost for each generator is expanded as

\displaystyle C_1=3x_1+\frac{1}{40}x_1^2

\displaystyle C_2=3x_2+\frac{2}{40}x_2^2

\displaystyle C_3=3x_3+\frac{3}{40}x_3^2

The total cost of production is

\displaystyle C(x_1,x_2,x_3)=3x_1+\frac{1}{40}x_1^2+3x_2+\frac{2}{40}x_2^2+3x_3+\frac{3}{40}x_3^2

Simplifying and rearranging, we have the objective function to minimize:

\displaystyle C(x_1,x_2,x_3)=3(x_1+x_2+x_3)+\frac{1}{40}(x_1^2+2x_2^2+3x_3^2)

The restriction can be modeled as a function g(x)=0:

g: x_1+x_2+x_3=1000

Or

g(x_1,x_2,x_3)= x_1+x_2+x_3-1000

We now construct the auxiliary function

f(x_1,x_2,x_3)=C(x_1,x_2,x_3)-\lambda g(x_1,x_2,x_3)

\displaystyle f(x_1,x_2,x_3)=3(x_1+x_2+x_3)+\frac{1}{40}(x_1^2+2x_2^2+3x_3^2)-\lambda (x_1+x_2+x_3-1000)

We find all the partial derivatives of f and equate them to 0

\displaystyle f_{x1}=3+\frac{2}{40}x_1-\lambda=0

\displaystyle f_{x2}=3+\frac{4}{40}x_2-\lambda=0

\displaystyle f_{x3}=3+\frac{6}{40}x_3-\lambda=0

f_\lambda=x_1+x_2+x_3-1000=0

Solving for \lambda in the three first equations, we have

\displaystyle \lambda=3+\frac{2}{40}x_1

\displaystyle \lambda=3+\frac{4}{40}x_2

\displaystyle \lambda=3+\frac{6}{40}x_3

Equating them, we find:

x_1=3x_3

\displaystyle x_2=\frac{3}{2}x_3

Replacing into the restriction (or the fourth derivative)

x_1+x_2+x_3-1000=0

\displaystyle 3x_3+\frac{3}{2}x_3+x_3-1000=0

\displaystyle \frac{11}{2}x_3=1000

x_3=181.8\ MW

And also

x_1=545.5\ MW

x_2=272.7\ MW

The load balance (x_1,x_2,x_3)=(545.5,272.7,181.8) Mw minimizes the total cost

5 0
3 years ago
Rewrite the expression by combining like terms.<br> 3xy + 8x^2+ 4y + 2y - 5x^2
Lelu [443]

Answer:

3x^2 + 3xy + 6y

Step-by-step explanation:

7 0
3 years ago
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