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inysia [295]
4 years ago
9

which geometric series represents 0.4444... as a fraction? a) 1/4, 1/40, 1/400, 1/4,000 b) 1/40, 1/400, 1/4,000, 1/40,000 c) 4/1

0, 4/100, 4/1,000, 4/10,00 d) 1/10, 1/100, 1/1000, 1/10000
Mathematics
2 answers:
xenn [34]4 years ago
7 0

Answer: \frac{4}{10}+\frac{4}{100}+\frac{4}{1,000}+\frac{4}{10,000}


Step-by-step explanation:

The given geometric series are:

a)\frac{1}{4}+\frac{1}{40}+\frac{1}{400}+\frac{1}{4,000}

on simplifying in decimals, we get

0.25+0.025+0.0025+0.00025=0.27775\neq0.4444

b) \frac{1}{40}+\frac{1}{400}+\frac{1}{4,000}+\frac{1}{40,000}

on simplifying in decimals, we get

0.025+0.0025+0.00025+0.000025=0.027775\neq0.4444

c)\frac{4}{10}+\frac{4}{100}+\frac{4}{1,000}+\frac{4}{10,000}

on simplifying in decimals, we get

0.4+0.04+0.004+0.0004=0.4444

Thus, this geometric series represent 0.4444.

d)\frac{1}{10}+\frac{1}{100}+\frac{1}{1,000}+\frac{1}{10,000}

on simplifying in decimals, we get

0.1+0.01+0.001+0.0001=0.1111\ \neq0.4444

viktelen [127]4 years ago
4 0
The answer is <span>c) 4/10, 4/100, 4/1,000, 4/10,00</span>
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The simplest way to find percent is to take the partial number, divide by the whole number, and multiply the answer by 100%.

= 0.4

0.4 x 100% = 40%.
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3 years ago
Three balls are selected at random without replacement from an urn containing two white balls and four blue balls. Find the prob
guapka [62]

<u>Answer:</u>

P(3 blue) = 1/5

<u>Steps:</u>

P(3 blue) = 2/3 × 3/5 × 1/2

P(3 blue) = 6/30

P(3 blue) = 1/5

<em>that's 20%</em>

4 0
3 years ago
Answer???????? with explanations, para lam ko di mga hula
Arisa [49]

Answer:

144

Step-by-step explanation:

<u>To </u><u>Find</u><u> </u><u>:</u><u>-</u><u> </u>

  • Least Common denominator .

<u>Solution</u><u> </u><u>:</u><u>-</u><u> </u>

We have ,

> 1/8 , 2/9 , 3/12 .

The denominator of the fractions are ,

> 8 , 9 , 12

The LCM of 8,9,12 will be ,

2 |<u> </u><u>8</u><u> </u><u>,</u><u> </u><u>9</u><u> </u><u>,</u><u> </u><u>1</u><u>2</u><u> </u>

2 <u>| 4 , 9 , 6</u>

2 |<u> </u><u>2</u><u> </u><u>,</u><u> </u><u>9</u><u> </u><u>,</u><u> </u><u>3 </u>

3 |<u> </u><u>2</u><u> </u><u>,</u><u> </u><u>3 </u><u>,</u><u> </u><u>1</u><u> </u>

Therefore , LCM will be ,

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7 0
3 years ago
In response to the increasing weight of airline passengers, the Federal Aviation Administration in 2003 told airlines to assume
kobusy [5.1K]

Answer:

The probability that the total weight of the passengers exceeds 4222 pounds is 0.0018

Step-by-step explanation:

The Central limit Theorem stays that for a large value of n (21 should be enough), the average distribution X has distribution approximately normal with mean equal to 182 and standard deviation equal to 30/√21 = 6.5465. Lets call W the standarization of X. W has distribution approximately N(0,1) and it is given by the formula

W = \frac{X - \mu}{\sigma} = \frac{X - 182}{6.5465}

In order for the total weight to exceed 4222 pounds, the average distribution should exceed 4222/21 = 201.0476.

The cummulative distribution function of W will be denoted by \phi . The values of \phi can be found in the attached file.

P(X > 201.0476) = P(\frac{X-182}{6.5465} > \frac{201.0476 - 182}{6.5465}) = P(W > 2.91) = 1-\phi(2.91) = \\1-0.9982 = 0.0018

Therefore, the probability that the total weight of the passengers exceeds 4222 pounds is 0.0018.  

Download pdf
6 0
4 years ago
Use multiplication to find 2 equivalent fraction for 3/4 and 4/5
Contact [7]
Equivalent fractions for 3/4 are 6/8 and 9/12

Equivalent fractions for 4/5 is 8/10 and 12/15
7 0
3 years ago
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