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FromTheMoon [43]
2 years ago
15

How many times larger is the value 250,000 then 250

Mathematics
2 answers:
Olegator [25]2 years ago
4 0
1,000 (it needs 20 characters)...
saveliy_v [14]2 years ago
3 0

Answer:

1,000 times larger.

Step-by-step explanation:

For these types of questions, simply divide the larger number by the smaller number. Like this:

250,000 ÷ 250 = 1,000

<em>Hope this helps!</em>

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An initial amount of $2000 is invested in an account at an interest rate of 5% per year, compounded continuously. Find the amoun
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Step-by-step explanation:

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X/72 = 135/90 what is x?
enot [183]
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5 0
3 years ago
A box contains 20 light box of which five or defective it for lightbulbs or pick from the box randomly what's the probability th
Snowcat [4.5K]

Answer:

1

Step-by-step explanation:

Given:-

- The box has n = 20 light-bulbs

- The number of defective bulbs, d = 5

Find:-

what's the probability that at most two of them are defective

Solution:-

- We will pick 2 bulbs randomly from the box. We need to find the probability that at-most 2 bulbs are defective.

- We will define random variable X : The number of defective bulbs picked.

Such that,               P ( X ≤ 2 ) is required!

- We are to make a choice " selection " of no defective light bulb is picked from the 2 bulbs pulled out of the box.

- The number of ways we choose 2 bulbs such that none of them is defective, out of 20 available choose the one that are not defective i.e n = 20 - 5 = 15 and from these pick r = 2:

        X = 0 ,       Number of choices = 15 C r = 15C2 = 105 ways

- The probability of selecting 2 non-defective bulbs:

      P ( X = 0 ) = number of choices with no defective / Total choices

                       = 105 / 20C2 = 105 / 190

                       = 0.5526

- The number of ways we choose 2 bulbs such that one of them is defective, out of 20 available choose the one that are not defective i.e n = 20 - 5 = 15 and from these pick r = 1 and out of defective n = 5 choose r = 1 defective bulb:

        X = 1 ,       Number of choices = 15 C 1 * 5 C 1 = 15*5 = 75 ways

- The probability of selecting 1 defective bulbs:

      P ( X = 1 ) = number of choices with 1 defective / Total choices

                       = 75 / 20C2 = 75 / 190

                       = 0.3947

- The number of ways we choose 2 bulbs such that both of them are defective, out of 5 available defective bulbs choose r = 2 defective.

        X = 2 ,       Number of choices = 5 C 2 = 10 ways

- The probability of selecting 2 defective bulbs:

      P ( X = 2 ) = number of choices with 2 defective / Total choices

                       = 10 / 20C2 = 10 / 190

                       = 0.05263

- Hence,

    P ( X ≤ 2 ) = P ( X =0 ) + P ( X = 1 ) + P (X =2)

                     = 0.5526 + 0.3947 + 0.05263

                     = 1

7 0
3 years ago
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