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stepladder [879]
1 year ago
14

Alex wrote the expanded form for the number 165.038 as "100 60 5 30 8." was he correct? if not, give the correct expanded form.

Mathematics
1 answer:
grin007 [14]1 year ago
5 0

The expanded form for the number 165.038 is 100+60+5+0.03+0.008.

<h3>What is the expended form of decimal numbers?</h3>

Writing decimals in expanded form simply means writing each number according to its place value. This is done by multiplying each digit by its place value and adding them together. Let's look at an example: 2.435. In words, we would say this as two and four hundred thirty-five thousandths.

In the given question expended form 100+60+5+30+8 is wrong.

After decimal point the first digit is at the tenth place and second digit at the hundredth place and third is at the thousandth place and so on....

165.038  = 100+60+5+\frac{0}{10}+\frac{3}{100}+\frac{8}{1000}            

              = 100+60+5+0.03+0.008

Hence, The correct expended form of the given number is 100+60+5+0.03+0.008.

To learn more about expended form from the given link:

brainly.com/question/603715

#SPJ4

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Answer:

Following are the solution to this question:

Step-by-step explanation:

Please find the complete question in the attached file.

In the given equation, when the point t=0

So,

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A random experiment was conducted where a Person A tossed five coins and recorded the number of ""heads"". Person B rolled two d
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Answer:

(10) Person B

(11) Person B

(12) P(5\ or\ 6) = 60\%

(13) Person B

Step-by-step explanation:

Given

Person A \to 5 coins (records the outcome of Heads)

Person \to Rolls 2 dice (recorded the larger number)

Person A

First, we list out the sample space of roll of 5 coins (It is too long, so I added it as an attachment)

Next, we list out all number of heads in each roll (sorted)

Head = \{5,4,4,4,4,4,3,3,3,3,3,3,3,3,3,3,2,2,2,2,2,2,2,2,2,2,1,1,1,1,1,0\}

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Next, we list out the highest in each toss (sorted)

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Question 10: Who is likely to get number 5

From person A list of outcomes, the proportion of 5 is:

Pr(5) = \frac{n(5)}{n(Head)}

Pr(5) = \frac{1}{32}

Pr(5) = 0.03125

From person B list of outcomes, the proportion of 5 is:

Pr(5) = \frac{n(5)}{n(Dice)}

Pr(5) = \frac{8}{30}

Pr(5) = 0.267

<em>From the above calculations: </em>0.267 > 0.03125<em> Hence, person B is more likely to get 5</em>

Question 11: Person with Higher median

For person A

Median = \frac{n(Head) + 1}{2}th

Median = \frac{32 + 1}{2}th

Median = \frac{33}{2}th

Median = 16.5th

This means that the median is the mean of the 16th and the 17th item

So,

Median = \frac{3+2}{2}

Median = \frac{5}{2}

Median = 2.5

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Median = \frac{n(Dice) + 1}{2}th

Median = \frac{30 + 1}{2}th

Median = \frac{31}{2}th

Median = 15.5th

This means that the median is the mean of the 15th and the 16th item. So,

Median = \frac{5+5}{2}

Median = \frac{10}{2}

Median = 5

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Question 12: Probability that B gets 5 or 6

This is calculated as:

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From the sample space of person B, we have:

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n(5\ or\ 6) = 18

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P(5\ or\ 6) = \frac{n(5\ or\ 6)}{n(Dice)}

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P(3\ or\ more) = \frac{n(3\ or\ more)}{n(Head)}

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P(3\ or\ more) = 50\%

Person B

n(3\ or\ more) = 28

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P(3\ or\ more) = \frac{n(3\ or\ more)}{n(Dice)}

P(3\ or\ more) = \frac{28}{30}

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P(3\ or\ more) = 93.3\%

By comparison:

93.3\% > 50\%

Hence, person B has a higher probability of 3 or more

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