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2 years ago

Am I correct with this?

Mathematics

1 answer:

Shkiper50 [21]2 years ago

7 0

Answer:

Step-by-step explanation:

u can cut the trapesoid in half and make it into a rectangle then get ur answer

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Building Bonds puts on team-building events. The company charges a base rate of $45 per event, plus $7.50 per participant. Write

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Answer:

45+7.50x= y

Step-by-step explanation:

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2 years ago

Identify the x-intercept

sveta [45]

im quite positive that the x-intercept is 1,6

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2 years ago

F of x equals 8 plus 2x eminus x squared

maria [59]

Answer:

Step-by-step explanation:

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2 years ago

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Use the set of data to calculate the measures that follow.

zepelin [54]

Answer:

<h3>Mean=3.5</h3><h3>Median=3.5</h3><h3>Range=6</h3><h3>Interquartile=1</h3>

Step-by-step explanation:

Given set of data is 0, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 6

<h3>To find Mean , Median , Range and Interquartile range :</h3>

First finding Mean

$Mean=\frac{sum of the observations}{number of observations}$
$=\frac{0+2+3+3+3+3+3+4+4+4+4+5+5+6}{14}$
$=\frac{49}{14}$
$=\frac{7}{2}$
$=3.5$

<h3>Therefore Mean=3.5</h3><h3>Median:</h3>

Since the number of observations is even, so the meadian becomes
$Median=\frac{sum of the two mid terms}{2}$
$=\frac{3+4}{2}$
$=\frac{7}{2}$
$=3.5$

<h3>Therefore Median=3.5</h3><h3>Range:</h3>

Range=greatest value-least value
In the given observations we have greatest value is 6 and least value is 0
Therefore Range=6-0

<h3>Therefore Range=6</h3><h3>Interquartile:</h3>

From the observations we have $Q_1=3$ and $Q_3=4$
$Interquartile=Q_3-Q_1$
$=4-3$

<h3>Therefore Interquartile=1</h3><h3 />

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2 years ago

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HELP PLEASE IF YOU DO THANKS!

denpristay [2]

Answer:

the pics are blocked

Step-by-step explanation:

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2 years ago