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3 years ago

What is the median of the data in this stem-and-leaf plot?

Mathematics

2 answers:

Gnom [1K]3 years ago

5 0

Answer:

Part A) Median = 44

Part B) Median = 36

Step-by-step explanation:

Median of a data set is the middle data point.

Part A) Median of the data in this stem-and-leaf plot.

The total data set represented by Stem and leaf in order from least to greatest are as following:

37 , 38 , 38 , 40 , 43 , 45 , 47 , 47 , 51 , 52

The middle data is 43 , 45

So, median = (43 + 45)/2 = 88/2 = 44

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Part B) Median of the data in A number line.

The total data set represented by the number line in order from least to greatest are as following:

31, 34 , 34 , 35 , 37 , 38 , 38 , 39

The middle data is 35 , 37

So, median = (35 + 37)/2 = 72/2 = 36

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zmey [24]3 years ago

5 0

Answer:

C) 44

Step-by-step explanation:

I took the test and it was right.

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A 5-foot piece of aluminum wire costs $7.80. What is the price per inch?

dmitriy555 [2]

Answer:

The answer would be .35 cent.

Step-by-step explanation:

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3 years ago

How to do this question plz

hjlf

Answer:

$\underline{ \boxed{ \underline{part \: a}}}. \\the \: required \: distance \: a part \: be \to \: \boxed{\underline {\overline{p } =79.23 \: km} } \\ \\ \underline{ \boxed{ \underline{part \: b}}}. \\ the \: bearing \: of \: \boxed{ B } \: \: from \: \boxed{ \underline{A}} \: is \to \boxed{ \underline{324 }\degree }$

Step-by-step explanation:

$\underline{ \boxed{ \underline{part \: a}}}. \\ l et \: the \: required \: distance \: apart \: be \to \: \boxed{ \overline {p}} \\ let \: \boxed{ \angle \: P} \: be \: the \: angle \: of \: seperation \: between \: both \: ships :given \: by \to \\ P = (244 - 180) - (196 - 180 )= (64 -16) = \boxed{48 \degree} \\ if \: ship \: A \: and \: B \: left \: the \: port\: at \: (10:30) \: \\ then \: at \: (14 :00) \: their \: time \: interval \: would \: be \to \\ (14 : 00) - (10 : 30) = \boxed{ 3.5 \: hrs} \\ ship \: \boxed{A }\: distance \: from \: the \: port =( v \times t) = (30 \times 3.5) = \boxed{105 \: km} \\ ship \: \boxed{B }\: distance \: from \: the \: port =( v \times t) = (24 \times 3.5) = \boxed{84\: km} \\n ow........we \: cant \: do \: much \: if \: their \: are \: no \: angles \: to \: work \: with \to \\ applying \:t he \: cosine \: rule : we \: have \to \\ \cos(P) = \frac{ {a}^{2} + {b}^{2} - \overline{{p}^{2} } }{2(ab)} \\ {a}^{2} + {b}^{2} - \overline{{p}^{2} } = 2(ab) \cos(P) \\ \overline{{p}^{2} } = {a}^{2} + {b}^{2} - \{2(ab) \cos(P) \} \\ \overline{{p}^{2} } = {84}^{2} + {105}^{2} - \{2(84)(105) \cos(48 \degree) \} \\\overline{{p}^{2} } = 18,081 - 11,803.463897 \\ \overline{{p}^{2} } = 6,277.536103 \\ \overline{p } = \sqrt{6,277.536103} \\ \boxed{\underline {\overline{p } =79.23 \: km} } \\ \\ \underline{ \boxed{ \underline{part \: b}}}. \\ to \: sove \: this : we \: first \: find \: angle \: \angle \: B : by \: applying \to \\ \frac{ \sin(B) }{b} = \frac{ \sin(P) }{p} \\ p \ast \sin(B) = b \ast \sin(P) \\ B = \sin {}^{ - 1} ( \frac{ b \ast \sin(P)}{p} ) \\ B = \sin {}^{ - 1} ( \frac{ 105 \times \sin(48 \degree)}{79.230903712} ) \\ \\ B = \sin {}^{ - 1} ( \frac{ 78.030206678}{79.230903712} ) \\ \\ B = \sin {}^{ - 1} (0.9848455971) \\ \boxed{B = 80 \degree} \\ hence \ : the \: bearing \: of \: \boxed{ B } \: \: from \: \boxed{ \underline{A}} \: is \to \\ 360 - \{180 - (80 + 48 + 16) \} \\ 360 -(180 - 144 )\\ 360 -36 = \boxed{324 \degree }$

♨Rage♨

7 0

3 years ago

Evaluation the following:

Angelina_Jolie [31]

Answer:

A. √32y²/72x²= 32y/72x

Step-by-step explanation:

√32y²/72x²= 32²/72²

= 32/72.

B. √125m³/36m²

=4192.6/108

=38.8203.

C. √x²/y²-y²/x²

therefore, x²=1, y²=-1

that is, √1/-1 - -1/1= -1(-)-1=0.

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Answer:

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