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gregori [183]
3 years ago
15

Which is the correct way to write three and one-tenth of a milliliter as an Arabic number?

Mathematics
1 answer:
Leya [2.2K]3 years ago
4 0
     we have that
<span>three and one-tenth------> 3 1/10--------> (3*10+1)/10--------> 31/10-----> 3.1 

the answer is 3.1 ml</span>
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Answer:7.25 feet

Step-by-step explanation:

6.25+6.5=12.75

20-12.75=7.25

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Step-by-step explanation:

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3 years ago
Eight less then 11 times a number is 47​
Studentka2010 [4]

Answer: X = 5

Step-by-step explanation:

The correct formula to solve the problem is:

11x - 8 = 47

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11x = 55

Solve it by dividing both sides by 11:

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8 0
3 years ago
8. There are 10 balls in a bag numbered from 1 to 10. Three balls are selected at random without replacement.
PIT_PIT [208]

Answer:

a. 720

b. 0.3

Step-by-step explanation:

a.

Three balls are selected at random without replacement.

First ball can be selected in 10 ways

Second ball can be selected in 9 ways

Third ball can be selected in 8 ways

Thus, there are 10×9×8 =720 different ways of selecting three balls.

b.

There are three cases that satisfy this condition

  1. first ball picked is 5, the others are not
  2. first ball is not 5, second is 5, third is not
  3. first and second ball is not five, third ball is 5

Each probabilities are:

  1. \frac{1}{10} * \frac{9}{9} *\frac{8}{8} =\frac{1}{10}
  2. \frac{9}{10} *\frac{1}{9} *\frac{8}{8} =\frac{1}{10}
  3. \frac{9}{10} *\frac{8}{9} *\frac{1}{8} =\frac{1}{10}

Thus, one of the balls selected is the number 5 is then \frac{1}{10} + \frac{1}{10} + \frac{1}{10} = \frac{3}{10}

5 0
3 years ago
Find the expansion of cos x about the point x=0
ycow [4]

Answer:

Cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{1!} + ...

Step-by-step explanation:

We use Taylor series expansion to answer this question.

We have to find the expansion of cos x at x = 0

f(x) = cos x, f'(x) = -sin x, f''(x) = -cos x, f'''(x) = sin x, f''''(x) = cos x

Now we evaluate them at x = 0.

f(0) = 1, f'(0) = 0, f''(0) = -1, f'''(0) = 0, f''''(0) = 1

Now, by Taylor series expansion we have

f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)(x-a)^2}{2!} + \frac{f'''(a)(x-a)^3}{3!} + \frac{f''''(a)(x-a)^4}{4!} + ...

Putting a = 0 and all the values from above in the expansion, we get,

Cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{1!} + ...

8 0
3 years ago
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