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Debora [2.8K]
3 years ago
15

a. Determine whether the ratios 10/20 and 30/60 can form a proportion by finding a common multiplier. b. Show that the ratios in

part (a) are equal by writing them in simplest form. Please explain how you got the answers to these questions.
Mathematics
1 answer:
Flura [38]3 years ago
6 0
<span>To simplify one by one, step by step, proving that they are the same:
</span>
\frac{10}{20} \frac{\div2}{\div2}  =  \frac{5}{10}  \frac{\div5}{\div5} =  \boxed{\frac{1}{2} } \end{array}}\qquad\quad\checkmark
\frac{30}{60} \frac{\div2}{\div2} =  \frac{15}{30}  \frac{\div5}{\div5} =  \frac{3}{6}  \frac{\div3}{\div3} =  \boxed{\frac{1}{2} } \end{array}}\qquad\quad\checkmark


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120

Step-by-step explanation:

4 0
3 years ago
Karen karlin bought some large frames for $15 each and some small frames for $8 each at a closeout sale. if she bought 22 frames
BaLLatris [955]

small frames ($8): s

large frames ($15): L


Cost:        8s + 15L = 239   ⇒ 1(8s + 15L = 239)   ⇒    8s + 15L = 239  

Quantity:   s  +   L   = 22   ⇒  -8( s  +   L   = 22)   ⇒  <u> -8s   -8 L  = -176 </u>

                                                                                               7L  = 63

                                                                                                  L = 9

Quantity:   s + L = 22 ⇒   s + (9) = 22   ⇒   s = 13

Answer: 13 small frames, 9 Large frames

7 0
3 years ago
Read 2 more answers
The following are the ages of 13 history teachers in a school district. 24, 27, 29, 29, 35, 39, 43, 45, 46, 49, 51, 51, 56 Notic
pishuonlain [190]

The five-number summary and the interquartile range for the data set are given as follows:

  • Minimum: 24.
  • Lower quartile: 29.
  • Median: 43.
  • Upper quartile: 50.
  • Maximum: 56.
  • Interquartile range: 50 - 29 = 21.

<h3>What are the median and the quartiles of a data-set?</h3>

  • The median of the data-set separates the bottom half from the upper half, that is, it is the 50th percentile.
  • The first quartile is the median of the first half of the data-set.
  • The third quartile is the median of the second half of the data-set.
  • The interquartile range is the difference between the third quartile and the first quartile.

In this problem, we have that:

  • The minimum value is the smallest value, of 24.
  • The maximum value is the smallest value, of 56.
  • Since the data-set has odd cardinality, the median is the middle element, that is, the 7th element, as (13 + 1)/2 = 7, hence the median is of 43.
  • The first quartile is the median of the six elements of the first half, that is, the mean of the third and fourth elements, mean of 29 and 29, hence 29.
  • The third quartile is the median of the six elements of the second half, that is, the mean of the third and fourth elements of the second half, mean of 49 and 51, hence 50.
  • The interquartile range is of 50 - 29 = 21.

More can be learned about five number summaries at brainly.com/question/17110151

#SPJ1

3 0
2 years ago
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