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Gala2k [10]
3 years ago
6

6 1⁄3 + 7 1⁄4 – 2 1⁄2 =

Mathematics
1 answer:
Korvikt [17]3 years ago
6 0

Answer:

11 1/12

Step-by-step explanation:

6 1/3 = 19/3

7 1/4 = 29/4

2 1/2 = 5/2

19/3 + 29/4 - 5/2


We must find the LCM of 3, 4, and 2. This happens to be 12

3*4 = 12

4*3 = 12

2*6 = 12

Multiply each fraction by the factor that'll get it to 12.


19/3 * 4/4 = 76/12

29/4 * 3/3 = 87/12

5/2 * 6/6 = 30/12

Now go through the problem


76/12 + 87/12 - 30/12

76 + 87 = 163

163/12 - 30/12

163 - 30 = 133

133/12

Simplify

133/12 = 11.0833... or 11 1/12


Hope this helps.

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Define the random variables X and X in words. X is the amount of time an individual waits at the courthouse to be called for ser
densk [106]

<u>Question</u>

Suppose that a committee is studying whether or not there is excessive time waste in our judicial system. It is interested in the mean amount of time individuals spend at the courthouse waiting to be called for jury duty. The committee randomly surveyed 81 people who recently served as jurors. The sample mean wait time was 4 hours with a sample standard deviation of 1.2 hours. Define the random variables X and \bar{X} in words.

Answer:

(B)

  • X is the amount of time an individual waits at the courthouse to be called for service.
  • \bar{X}  is the mean wait time for a sample of individuals.

Step-by-step explanation:

The random variable X is the amount of time for which each of the prospective juror waits at the courthouse before being called for service.

\bar{X}  is the mean wait time for a the given sample of individuals. In the case given,

5 0
2 years ago
The areas of two similar octagons are 9 m² and 25 m². What is the scale factor of their side lengths? PLZ HELP PLZ PLZ PLZ
torisob [31]

Answer:

\frac{3}{5}

Step-by-step explanation:

Let the side length for the octagon having 9m² as area = x

Side length for the octagon having area of 25m² = y.

Thus:

\frac{9}{25} = (\frac{x}{y})^2 (area of similar polygons theorem)

The scale factor of their sides would be \frac{x}{y}. Which is:

\sqrt{\frac{9}{25}} = \frac{x}{y}

\frac{\sqrt{9}}{\sqrt{25}} = \frac{x}{y}

\frac{3}{5} = \frac{x}{y}

Scale factor of their sides = \frac{3}{5}

6 0
2 years ago
List elements in..<br> {x | x is an even counting number smaller than 8}
ohaa [14]

Answer:

x l 0 2 4 6

Step-by-step explanation:

7 0
3 years ago
Which number would make the sentence true?
AURORKA [14]

Answer:

22% = 0.22

Step-by-step explanation:

Lat'e express all the given numbers in decimal form so we understand how they are located on the number line.

We start with 2/11 , since the number we are looking for has to be larger than this \, and smaller than 1.42:

\frac{2}{11} = 0.181818181818...

So we understand that we are looking for a number that can be placed between the lower boundary (0.181818...) and the upper boundary (1.42) on the number line. That is: we are looking for a number greater than 0.18181818... and less than 1.42.

Now let's look at each of the given options (also writing them in decimal form to facilitate the comparison with the lower and upper boundaries we just found):

1\frac{2}{3} = 1.66666666... This number is grater than the upper boundary given to us (1.42), it would be placed to the right of 1.42 on the number line. Therefore we discard it for not being in the requested interval (section) of the number line.

0.153 is already in decimal form, and clearly less than (thus to be placed on the left of) the lower boundary (0.181818...) of the requested interval. Therefore we discard it for not being in the requested interval (section) of the number line.

22% in decimal form is written as: \frac{22}{100} = 0.22. This number is greater than the lower boundary (0.181818...) and also less than the upper boundary (1.42). Therefore it is a choice that would make the sentence true.

1.3*10^1=1.3*10=13 which is clearly greater than the upper boundary of the interval, so we discard it.

7 0
3 years ago
Read 2 more answers
Want Brainliest?
andrew11 [14]

Answer:

6 hours  

Step-by-step explanation:

3 0
2 years ago
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