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2 years ago

A square has sides measuring 3 feet. If the sides of a square are doubled, will the area also double? Explain

1 answer:

8 0

Hello!

Let's find the area and see!

A = s^2

A = 3^2

A = 6 ft^2

Now, if you double 3, you get 6. 3 × 2 = 6 Find the area:

A = 6^6

A = 36 ft^2

ANSWER:

No, the area will not be doubled. The area was 9 ft^2 when the sides were 3 feet long. If you double 9, you get 18, since 9 × 2 = 18.

36 is not the double of 9, so no, the area is not double.

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Which fraction represents the decimal 0. ModifyingAbove 1 2 with Bar? One-twelfth StartFraction 3 Over 25 EndFraction StartFract

mestny [16]

The considered decimal number is expressed by the fraction given by: Option C: 4/33

<h3>How to convert from decimal to fraction?</h3>

For conversion from decimal to fraction, we write it in the form a/b such that the result of the fraction comes as the given decimal. Usually, to get the decimal of the form a.bcd, we count how many digits are there after the decimal point (if its finite), then we write 10 raised to that many power as denominator, and the considered number without any decimal point as numerator.

For example:

$0.99 = \dfrac{99}{10^2} = \dfrac{99}{100}$

(10 raised to power k = 1 followed by k zeroes)

<h3>What is the sum of an infinite geometric series?</h3>

The sum of an infinite geometric series having the initial term "a" and the multiplication per term is done by "r" such that r < 1

is given by

$S_{\infty} = \dfrac{a}{1-r}$

The series would look like $a+ ar+ ar^2+\cdots$

For the considered case, the given decimal number is

$0.\overline{12} = 0.121212\cdots$

We can check all the options available one by one, so as to know which one ends up giving this pattern, or we can use another technique as shown below.

$0.12121... = (0.1 + 0.001 + 0.00001 + ... ) + (0.02 + 0.0002 + 0.000002 + ... )$

Each of the two series obtained are geometric series with the multiplication factor as 1/100

Using the sum formula for evaluation of an infinite geometric series, and the fact that initial term of first series is 0.1 and that of second series is 0.02, we get:
$0.121212... = \dfrac{0.1}{1-1/100} + \dfrac{0.02}{1 - 1/100} = \dfrac{1}{99/100}(\dfrac{1}{10} + \dfrac{2}{100})\\0.121212... = \dfrac{100}{99}\times(\dfrac{10}{100} + \dfrac{2}{100})\\\\\\0.121212 ... = \dfrac{12}{99} = \dfrac{4}{33}$

Thus, the considered decimal number is expressed by the fraction given by: Option C: 4/33

Learn more about geometric series here:

brainly.com/question/16037289

4 0

1 year ago

4 + (y + 3) + y2 for y = 5

sergiy2304 [10]

Answer: 22

Step-by-step explanation:

just plug in 5 Hope I help u

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2 years ago

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Two teams A and B play a series of games until one team wins three games. We assume that the games are played independently and

Olenka [21]

Answer:

The probability that the series lasts exactly four games is $3p(1-p)(p^{2} + (1 - p)^{2})$

Step-by-step explanation:

For each game, there are only two possible outcomes. Either team A wins, or team A loses. Games are played independently. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

We also need to know a small concept of independent events.

Independent events:

If two events, A and B, are independent, we have that:

$P(A \cap B) = P(A)*P(B)$

What is the probability that the series lasts exactly four games?

This happens if A wins in 4 games of B wins in 4 games.

Probability of A winning in exactly four games:

In the first two games, A must win 2 of them. Also, A must win the fourth game. So, two independent events:

Event A: A wins two of the first three games.

Event B: A wins the fourth game.

P(A):

A wins any game with probability p. 3 games, so n = 3. We have to find P(A) = P(X = 2).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(A) = P(X = 2) = C_{3,2}.p^{2}.(1-p)^{1} = 3p^{2}(1-p)$

P(B):

The probability that A wins any game is p, so P(B) = p.

Probability that A wins in 4:

A and B are independent, so:

$P(A4) = P(A)*P(B) = 3p^{2}(1-p)*p = 3p^{3}(1-p)$

Probability of B winning in exactly four games:

In the first three games, A must win one and B must win 2. The fourth game must be won by 2. So

Event A: A wins one of the first three.

Event B: B wins the fourth game.

P(A)

P(X = 1).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(A) = P(X = 1) = C_{3,1}.p^{1}.(1-p)^{2} = 3p(1-p)^{2}$

P(B)

B wins each game with probability 1 - p, do P(B) = 1 - p.

Probability that B wins in 4:

A and B are independent, so:

$P(B4) = P(A)*P(B) = 3p(1-p)^{2}*(1-p) = 3p(1-p)^{3}$

Probability that the series lasts exactly four games:

$p = P(A4) + P(B4) = 3p^{3}(1-p) + 3p(1-p)^{3} = 3p(1-p)(p^{2} + (1 - p)^{2})$

The probability that the series lasts exactly four games is $3p(1-p)(p^{2} + (1 - p)^{2})$

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3 years ago

A set of numbers is shown below:

bonufazy [111]

The answer is

A. {1, 3, 6}

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Answer:

8/x

Step-by-step explanation:

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2 years ago