All categories

3 years ago

What is another name for proportional relationships?

Mathematics

1 answer:

arsen [322]3 years ago

4 0

Answer:

equivalent i think

Step-by-step explanation:

You might be interested in

The Environmental Protection Agency records data on the fuel economy of many different makes of cars. Data on the mileage of 20

artcher [175]

Answer:

A) 8.5 miles per gallon

Step-by-step explanation:

Given

$x = \{12\ 13\ 15\ 16\ 16\ 17\ 18\ 18\ 19\ 19\ 20\ 20\ 22\ 23\ 24\ 26\ 26\ 27\ 27\ 29\}$

$n = 20$

Required

The IQR

This is calculated as:

$IQR = Q_3 - Q_1$

Calculate $Q_1$

$Q_1 = \frac{1}{4}(n + 1)th$

$Q_1 = \frac{1}{4}(20 + 1)th$

$Q_1 = \frac{1}{4}(21)th$

Remove bracket

$Q_1 = 5.25th$

This means that:

$Q_1$ is the mean of 5th and 6th item

From the given data:

$5th \to 16$

$6th \to 17$

So, we have:

$Q_1 = \frac{1}{2}(16 + 17)$

$Q_1 = \frac{1}{2}(33)$

$Q_1 = 16.5$

$Q_1 = 16.25$

Calculate $Q_3$

$Q_3 = \frac{3}{4}(n + 1)th$

$Q_3 = \frac{3}{4}(20 + 1)th$

$Q_3 = \frac{3}{4}(21)th$

Remove bracket

$Q_3 = 15.75th$

This means that:

$Q_3$ is the mean of 15th and 16th item

From the data:

$15th \to 24$

$16th \to 26$

So:

$Q_3 = \frac{1}{2}(24 + 26)$

$Q_3 = \frac{1}{2}(50)$

$Q_3 = 25$

Recall that:

$IQR = Q_3 - Q_1$

$IQR = 25 -16.5$

$IQR = 8.5$

7 0

3 years ago

What is 5x-9=7x+6-2x

Sliva [168]

-9 = -6 is your answer

4 0

3 years ago

HELP MEEEEEEEEEE WILL MARK BRAINLESTI HAVE OVERDUE ASSIGNMENTS

ki77a [65]

Answer:

Step-by-step explanation:

2x - 7 = 27

2x = 34

x = 17

8 0

4 years ago

Pls solve this question

just olya [345]

Answer:

${( \sqrt{ {x}^{ - 3} }) }^{5} = ({( {x}^{ - 3}) }^{ \frac{1}{2} } ) ^{5} \\ \\ = {x}^{ - 3 \times \frac{1}{2} \times 5} = {x}^{ - \frac{15}{2} } = \frac{1}{ {x}^{ \frac{15}{2} } }$

<em><u>Or ;</u></em>

${( \sqrt{ {x}^{ - 3} } )}^{5} = {( \sqrt{ \frac{1}{ {x}^{3} }} })^{5} = ( { \frac{ \sqrt{1} }{ \sqrt{ {x}^{3} } } })^{5} \\ \\ = ( { \frac{1}{ \sqrt{x} \sqrt{ {x}^{2} } } })^{5} = ({ \frac{1}{ x\sqrt{x} }})^{5} \\ \\ = ( \frac{ {(1)}^{5} }{ {x}^{5} ( \sqrt{x} )^{5} } ) = \frac{1}{ {x}^{5} \times {x}^{ \frac{1}{2} \times 5 } } \\ \\ = \frac{1}{ {x}^{5} \times {x}^{ \frac{5}{2} } } = \frac{1}{ {x}^{5} \sqrt{ {x}^{5} } } \\ \\ = \frac{1}{ {x}^{5} \sqrt{ {x}^{4} {x}^{1} } } = \frac{1}{ {x}^{5} \sqrt{x} \sqrt{( { {x}^{2} })^{2} } } \\ \\ = \frac{1}{ {x}^{5} {x}^{2} \sqrt{x} } = \frac{1}{ {x}^{7} \sqrt{x} } = \frac{ \sqrt{x} }{ {x}^{8} }$

3 0

2 years ago

Your teacher for the discussion-based assessment.

vova2212 [387]

Answer:

Step-by-step explanation:

8 0

3 years ago