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3 years ago

What would 24.8 with the 8 repeating as an improper fraction

Mathematics

2 answers:

oksian1 [2.3K]3 years ago

8 0

31111111/25000 thats what i got

mojhsa [17]3 years ago

7 0

24.8 repeating is equivalent to $\frac{224}{9}$ in improper fractional form.

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A manufacturer ships its product in boxes with

Alla [95]

3 would be the volume hopes this is good

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3 years ago

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Which of these points is farthest from the origin?

Anastasy [175]

Answer:

d:-12,0

Step-by-step explanation:

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3 years ago

Miranda enlarged a picture twice as shown below, each time using a scale factor of 3.

lyudmila [28]

Answer:

The area of the second enlargement is 1,944 square inches

The area of the second enlargement is (3 squared) squared times the original area.

The ratio of the area of the first enlargement to the area of the original equals the square of the scale factor

Step-by-step explanation:

Verify each statement

1) The area of the first enlargement is 72 square inches.

The statement is false

Because

we know that

The original dimensions of the rectangle are

length 6 inches and width 4 inches

First enlargement

Multiply the original dimensions by a scale factor of 3

$Length: 6(3)=18\ inches\\Width: 4(3)=12\ inches$

The area of the first enlargement is

$18(12)=216\ in^2$

2) The area of the second enlargement is 1,944 square inches

The statement is true

Multiply the dimensions of the first enlargement by a scale factor of 3

$Length: 18(3)=54\ inches\\Width: 12(3)=36\ inches$

The area of the second enlargement is

$54(36)=1,944\ in^2$

3) The area of the second enlargement is (3 squared) squared times the original area.

The statement is true

Because

The original area is 24 square inches

$[(3^2)]^2(24)=1,944\ in^2$

4) The area of the second enlargement is 3 times the area of the first enlargement

The statement is false

Because

$3(216)=648\ in^2$

$648\ in^2 \neq 1,944\ in^2$

5) The ratio of the area of the first enlargement to the area of the original equals the square of the scale factor

The statement is true

Because

The square of the scale factor is 3^2=9

and the ratio is equal to

$\frac{216}{24}=9$

8 0

3 years ago

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WILL GIVE BRAINLIEST, THANKS AND FIVE STARS FOR CORRECT ANSWER

Kitty [74]

It’s twelve because you multiple x by 12 and get y.

4 0

3 years ago

Coach Evans recorded the height, in inches of each player on his team. The results are shown.

marysya [2.9K]

Answer:

Step-by-step explanation:

Given:

Team heights (inches):

61, 57, 63, 62, 60, 64, 60, 62, 63

To find: IQRs (interquartile ranges) of the heights for the team

Solution:

A quartile divides the number of terms in the data into four more or less equal parts that is quarters.

For a set of data, a number for which 25% of the data is less than that number is known as the first quartile $(Q_1)$

For a set of data, a number for which 75% of the data is less than that number is known as the third quartile $(Q_3)$

Terms in arranged in ascending order:

$57,60,60,61,62,62,63,63,64$

Number of terms = 9

As number of terms is odd, exclude the middle term that is 62.

$Q_1$ is median of terms $57,60,60,61$

Number of terms (n) = 4

Median = $\frac{(\frac{n}{2})^{th} +(\frac{n}{2}+1)^{th} }{2} =\frac{2^{nd}+3^{rd}}{2} =\frac{60+60}{2}=\frac{120}{2}=60$

So, $Q_1=60$

So, 25% of the heights of a team is less than 60 inches

$Q_3$ is the median of terms $62,63,63,64$

Median = $\frac{(\frac{n}{2})^{th} +(\frac{n}{2}+1)^{th} }{2} =\frac{2^{nd}+3^{rd}}{2} =\frac{63+63}{2}=\frac{126}{2}=63$

So, $Q_3=63$

So, 75% of the heights of a team is less than 63 inches

Interquartile range = $Q_3-Q_1=63-60=3$

The interquartile range is a measure of variability on dividing a data set into quartiles.

The interquartile range is the range of the middle 50% of the terms in the data.

So, 3 is the range of the middle 50% of the heights of the students.

4 0

3 years ago