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shusha [124]
3 years ago
10

2x+3y 26 3x +2y 6 Any solutions

Mathematics
1 answer:
Iteru [2.4K]3 years ago
8 0
6 might be helpful
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Evaluate the expression 5x4+2x6
vodka [1.7K]

Answer:

The answer to this is: 132

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Please answer this
pentagon [3]

Answer:

14 in^3.

Step-by-step explanation:

The volume = volume of the bottom prism + volume of the top pyramid

= area of base * height of the prism + 1/3 * area of base * height of the pyramid

= 2 * 1.5 * 4 + 1/3 * (2*1.5) * 2

= 3*4 + 1/3 * 6

= 14 in^3.

6 0
3 years ago
Multiply the equation 2x+2y=8 by 3. Does the equation have the same solution set?
stepan [7]

Multiply the equation:

2x+2y=8 \mapsto 6x+6y=24

The solution set is the same, because multiplying both sides of an equation by a non-zero number doesn't change the solution set. In fact, if you rewrite the equation as

2x+2y-8=0

Multiplying this by 3 (or whatever number, for all it matters) gives

3(2x+2y-8)=0

Now, a product is zero if and only if at least one of the factor is zero. So, either 3=0 or 2x+2y-8=0

Since the first is clearly impossible, the second one must be true, which is the original equation.

3 0
3 years ago
In a string of 12 Christmas tree light bulbs, 3 are defective. The bulbs are selected at random and tested, one at a time, until
Juliette [100K]

Using the hypergeometric distribution, it is found that there is a 0.0273 = 2.73% probability that the third defective bulb is the fifth bulb tested.

In this problem, the bulbs are chosen without replacement, hence the <em>hypergeometric distribution</em> is used to solve this question.

<h3>What is the hypergeometric distribution formula?</h3>

The formula is:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • N is the size of the population.
  • n is the size of the sample.
  • k is the total number of desired outcomes.

In this problem:

  • There are 12 bulbs, hence N = 12.
  • 3 are defective, hence k = 3.

The third defective bulb is the fifth bulb if:

  • Two of the first 4 bulbs are defective, which is P(X = 2) when n = 4.
  • The fifth is defective, with probability of 1/8, as of the eight remaining bulbs, one will be defective.

Hence:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}

P(X = 2) = h(2,12,4,3) = \frac{C_{3,2}C_{9,1}}{C_{12,4}} = 0.2182

0.2182 x 1/8 = 0.0273.

0.0273 = 2.73% probability that the third defective bulb is the fifth bulb tested.

More can be learned about the hypergeometric distribution at brainly.com/question/24826394

8 0
2 years ago
How do I solve number 10? I'm not sure if I did it right.
grandymaker [24]
No, you did it right
3 0
3 years ago
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