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Stella [2.4K]
3 years ago
6

What do you do to check whether a number is rational or irrational? In your explanation, use an example of an irrational and rat

ional number.
Can someone check my answer, please? I would deeply appreciate it.

"Rational numbers are numbers that can be written in fraction form. The numbers are all integers. Irrational numbers are numbers that cannot be written as a fraction. They are non-repeating and nonterminating decimals. For example 6.12385546972578945213666479952145412639854845256987452874526874125874........ Is an irrational number. And 1/6, 2,9, and 3/4 are all rational numbers."
Mathematics
1 answer:
nordsb [41]3 years ago
4 0
You have a good response. The only critique I can think of is to reduce the size of that massive decimal number. It doesn't have to be huge as long as it's a random string of digits. For example, e = 2.71828182846... is irrational since it has no pattern in the decimal portion. The decimal portion also goes on forever. 
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When you add two rational numbers, each number can be written as a?
ra1l [238]
<span>The correct answer to your question is... a rational #, because w</span><span>hen you add two rational #'s, each # can be written as a rational #. 
</span><span>
Reasoning:
So, adding two rational #'s like adding fractions will result in another fraction of this same form since integers are closed under + and x. Thus, adding two rational #'s produces another rational #.

By the way # means number.

</span>I hope this helps!
Please Rate & Thank!
Please mark as Brainliest!

Have a wonderful day! : )
8 0
3 years ago
Calculate the area of the following figure. All measurements are in centimetres. Give answer to correct two decimal places.
kykrilka [37]

Answer:

Step-by-step explanation:

The shape above is a composite shape having two different shapes like rectangle and semi- circle

Area of rectangle=Length × Breadth

Area of semi circle=1/2πr^2

To find the area of the rectangle naw

Length of the rectangle=6.6

Width of the rectangle=4

A=6.6×4

A=26cm^2

Area of semi-circle=1/2

Diameter of the shape =4cm

Radius=2cm

A=1/2×3.14×2cm x2cm

A=6.28cm^2

Area of the shape=26+6.28cm^2

Area=32.28 to 2 dp

3 0
3 years ago
X-26=4 add what to each side
Helen [10]
Add 26 to each side of the equation !
6 0
3 years ago
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Convert 3x-4y = 9 to slope intercept form. What are the slope and y - intercept ?
Alecsey [184]

Answer:

Slope intercept form; -4y = -3x + 9

Y-intercept; 9

3 0
3 years ago
Read 2 more answers
A random experiment was conducted where a Person A tossed five coins and recorded the number of ""heads"". Person B rolled two d
cestrela7 [59]

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\}

n(Head) = 32

Person B

First, we list out the sample space of toss of 2 coins (It is too long, so I added it as an attachment)

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

Dice = \{2,2,3,3,3,3,4,4,4,4,4,4,5,5,5,5,5,5,5,5,6,6,6,6,6,6,6,6,6,6\}

n(Dice) = 30

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

For person B

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

<em>Person B has a greater median of 5</em>

Question 12: Probability that B gets 5 or 6

This is calculated as:

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

From the sample space of person B, we have:

n(5\ or\ 6) =n(5) + n(6)

n(5\ or\ 6) =8+10

n(5\ or\ 6) = 18

So, we have:

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

P(5\ or\ 6) = \frac{18}{30}

P(5\ or\ 6) = 0.60

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

Question 13: Person with higher probability of 3 or more

Person A

n(3\ or\ more) = 16

So:

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

P(3\ or\ more) = \frac{16}{32}

P(3\ or\ more) = 0.50

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

Person B

n(3\ or\ more) = 28

So:

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

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

P(3\ or\ more) = 0.933

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

By comparison:

93.3\% > 50\%

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

7 0
3 years ago
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