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3 years ago

which one of the following examples represents a repeating decimal A. 1.111114 B. 0.777777 C. 4.252525 D. 0.123123

1 answer:

7 0

Actually, i think something is missing here:

You need either a parenthesis or some dots at the end to determine this. A repeating decimal can have one repreating digit:

0.(7): 0.777777...

two:

0.(45): 0.45454545454545....

or more: so potentially all of them can be repeating, even a!

it could be: 1.(111114)

or: 1.111114111114111114111114111114111114111114111114111114111114111114111114111114...

proably B. is the most typical of repeating decimals (choosed this one if you have to), but in reality, you need more information... did you copy the question exactly?

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How can you make your answer (41-13i)/(25) into a+bi form?

KengaRu [80]

Hello, do you know that for any a, b and c real numbers with c different from 0 the following is correct?

$\dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}$

Then,

$\dfrac{41-13i}{25}=\dfrac{41}{25}-\dfrac{13}{25}i$

So,

$a=\dfrac{41}{25}\\\\b=\dfrac{-13}{25}$

Thank you.

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3 years ago

Read 2 more answers

The numbers 1, 2, 3, 4, and 5 are written on slips of paper, and 2 slips are drawn at random one at a time without replacemen

mamaluj [8]

<h2>Answer:</h2>

(a)

The probability is : 1/2

(b)

The probability is : 1/2

<h2>Step-by-step explanation:</h2>

The numbers 1, 2, 3, 4, and 5 are written on slips of paper, and 2 slips are drawn at random one at a time without replacement.

The total combinations that are possible are:

(1,2) (1,3) (1,4) (1,5)

(2,1) (2,3) (2,4) (2,5)

(3,1) (3,2) (3,4) (3,5)

(4,1) (4,2) (4,3) (4,5)

(5,1) (5,2) (5,3) (5,4)

i.e. the total outcomes are : 20

(a)

Let A denote the event that the first number is 4.

and B denote the event that the sum is: 9.

Let P denote the probability of an event.

We are asked to find:

P(A|B)

We know that it could be calculated by using the formula:

$P(A|B)=\dfrac{P(A\bigcap B)}{P(B)}$

Hence, based on the data we have:

$P(A\bigcap B)=\dfrac{1}{20}$

( Since, out of a total of 20 outcomes there is just one outcome which comes in A∩B and it is: (4,5) )

and

$P(B)=\dfrac{2}{20}$

( since, there are just two outcomes such that the sum is: 9

(4,5) and (5,4) )

Hence, we have:

$P(A|B)=\dfrac{\dfrac{1}{20}}{\dfrac{2}{20}}\\\\i.e.\\\\P(A|B)=\dfrac{1}{2}$

(b)

Let A denote the event that the first number is 3.

and B denote the event that the sum is: 8.

Let P denote the probability of an event.

We are asked to find:

P(A|B)

Hence, based on the data we have:

$P(A\bigcap B)=\dfrac{1}{20}$

( since, the only outcome out of 20 outcomes is: (3,5) )

and

$P(B)=\dfrac{2}{20}$

( since, there are just two outcomes such that the sum is: 8

(3,5) and (5,3) )

Hence, we have:

$P(A|B)=\dfrac{\dfrac{1}{20}}{\dfrac{2}{20}}\\\\i.e.\\\\P(A|B)=\dfrac{1}{2}$

7 0

3 years ago