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sergejj [24]
3 years ago
6

Help me please!! first write the subtraction with a common denominator then subtract

Mathematics
2 answers:
Ivahew [28]3 years ago
6 0

Answer:

\frac{3}{5} - \frac{1}{3} = \frac{9}{15} - \frac{5}{15} = \frac4{15}

Step-by-step explanation:

3/5 - 1/3

Multiply 3/5 by 3/3 to keep the fraction equivalent and have a common denominator

3/5 times 3/3 = 9/15

Multiply 1/3 by 5/5 to keep the fraction equivalent and have a common denominator

1/3 times 5/5 = 5/15

9 / 15 - 5 / 15 = 4/15

If my answer is incorrect, pls correct me!

If you like my answer and explanation, mark me as brainliest!

-Chetan K

tatiyna3 years ago
3 0

Answer:

hope it helps you

Step-by-step explanation:

pls kindly mark it brainliest answer

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A DVD that originally cost $30 is on sale for 10 percent off. Complete the steps to solve the problem
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Karolina [17]

Answer:

1/2 or 50%

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Below are survival times (in days) of 13 guinea pigs that were injected with a bacterial infection in a medical study:
tresset_1 [31]

Outliers are data that are relatively far from other data elements.

The dataset has an outlier and the outlier is 120

The dataset is given as:

  • 91 83 84 79 91 93 95 97 97 120 101 105 98

Sort the dataset in ascending order

  • 79 83 84 91 91 93 95 97 97 98 101 105 120

<h3>The lower quartile (Q1)</h3>

The Q1 is then calculated as:

Q1 = \frac{N +1}{4}th

So, we have:

Q1 = \frac{13 +1}{4}th

Q1 = \frac{14}{4}th

Q1 = 3.5th

This is the average of the 3rd and the 4th element

Q1 = \frac{1}{2} \times (84 + 91)

Q1 = 87.5

<h3>The upper quartile (Q3)</h3>

The Q3 is then calculated as:

Q3 =  3 \times \frac{N +1}{4}th

So, we have:

Q3 =  3 \times \frac{13 +1}{4}th

Q3 =  3 \times 3.5th

Q3 =  10.5th

This is the average of the 10th and the 11th element.

Q_3 =\frac12 \times (98 + 101)

Q_3 =99.5

<h3>The interquartile range (IQR)</h3>

The IQR is then calculated as:

IQR = Q_3 -Q_1

IQR = 99.5 - 87.5

IQR = 12

Also, we have:

IQR(1.5) = 12 \times 1.5

IQR(1.5) = 18

<h3>The outlier range</h3>

The lower and the upper outlier range are calculated as follows:

Lower = Q_1 - IQR(1.5)

Lower = 87.5- 18

Lower = 69.5

Upper = Q_3 + IQR(1.5)

Upper = 99.5 + 18

Upper = 117.5

120 is greater than 117.5.

Hence, the dataset has an outlier and the outlier is 120

Read more about outliers at:

brainly.com/question/9933184

6 0
2 years ago
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