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Mekhanik [1.2K]
3 years ago
10

Please answer the 2 questions if you can? (This is my math from 8th grade)

Mathematics
1 answer:
LUCKY_DIMON [66]3 years ago
4 0

Answer:

Hi myself Shrushtee.

Step-by-step explanation:

8) a) x = 8.

10) c) x = 3

please mark me as brainleist

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The price of a pair of shoes increases from $55 to $77. What is the percent increase to the nearest percent?
Rom4ik [11]

Answer:

40%

Step-by-step explanation:

40% of 55 is 22 and 55+22=77

hoped this helped :)

6 0
3 years ago
Please explain how do you get the answer. Also tell me am I correct or wrong.
Zigmanuir [339]
Your doing everything correctly 
3 0
4 years ago
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for every 30 minutes of TV air time there are 8 minutes of commercial. If 90 minutes of TV or are now how many how many minutes
Anna35 [415]

Let m = minutes of commercial


30/90 = 8/m


Solve for m.


90(8) = 30m


720 = 30m


720/30 = m


24 = m

7 0
3 years ago
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I need help again...
Vadim26 [7]

the answer is 2 what you do to the bottom you do to the top

6 0
2 years ago
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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
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