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3 years ago

2.

Mathematics

2 answers:

777dan777 [17]3 years ago

7 0

Your answer choice should be ( B )

Darina [25.2K]3 years ago

5 0

Answer: B

0.07274

Step-by-step explanation:

Length

72.74

Millimeter

0.07274

Meter

Formula

divide the length value by 1000

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What is the difference between 1.37 and 0.52

Oksanka [162]

Answer:

0.85

Step-by-step explanation:

1.37-0.52=0.85

8 0

2 years ago

Two soccer teams play 8 games in their season. The number of goals each team scored per game is listed below: Team X: 11, 3, 0,

Gre4nikov [31]

Answer:

C. Team Y’s scores have a lower mean value.

Step-by-step explanation:

We are given that Two soccer teams play 8 games in their season. The number of goals each team scored per game is listed below:

Team X: 11, 3, 0, 0, 2, 0, 6, 4

Team Y: 4, 2, 0, 3, 2, 1, 6, 4

Firstly, we will calculate the mean, median, range and inter-quartile range for Team X;

Mean of Team X data is given by the following formula;

Mean, $\bar X$ = $\frac{\sum X}{n}$

= $\frac{11+ 3+ 0+ 0+ 2+ 0+ 6+ 4}{8}$ = $\frac{26}{8}$ = 3.25

So, the mean of Team X's scores is 3.25.

Now, for calculating the median; we have to arrange the data in ascending order and then observe that the number of observations (n) in the data is even or odd.

Team X: 0, 0, 0, 2, 3, 4, 6, 11

If n is odd, then the formula for calculating median is given by;

Median = $(\frac{n+1}{2} )^{th} \text{ obs.}$

If n is even, then the formula for calculating median is given by;

Median = $\frac{(\frac{n}{2})^{th} \text{ obs.} +(\frac{n}{2}+1)^{th} \text{ obs.} }{2}$

Here, the number of observations is even, i.e. n = 8.

So, Median = $\frac{(\frac{n}{2})^{th} \text{ obs.} +(\frac{n}{2}+1)^{th} \text{ obs.} }{2}$

= $\frac{(\frac{8}{2})^{th} \text{ obs.} +(\frac{8}{2}+1)^{th} \text{ obs.} }{2}$

= $\frac{(4)^{th} \text{ obs.} +(5)^{th} \text{ obs.} }{2}$

= $\frac{2+3}{2}$ = 2.5

So, the median of Team X's score is 2.5.

Now, the range is calculated as the difference between the highest and the lowest value in our data.

Range = Highest value - Lowest value

= 11 - 0 = 11

So, the range of Team X's score is 11.

Now, the inter-quartile range of the data is given by;

Inter-quartile range = $Q_3-Q_1$

$Q_1=(\frac{n+1}{4} )^{th} \text{ obs.}$

= $(\frac{8+1}{4} )^{th} \text{ obs.}$

= $(2.25 )^{th} \text{ obs.}$

$Q_1 = 2^{nd} \text{ obs.} + 0.25[ 3^{rd} \text{ obs.} -2^{nd} \text{ obs.} ]$

= 0 + 0.25[0 - 0] = 0

$Q_3=3(\frac{n+1}{4} )^{th} \text{ obs.}$

= $3(\frac{8+1}{4} )^{th} \text{ obs.}$

= $(6.75 )^{th} \text{ obs.}$

$Q_3 = 6^{th} \text{ obs.} + 0.75[ 7^{th} \text{ obs.} -6^{th} \text{ obs.} ]$

= 4 + 0.75[6 - 4] = 5.5

So, the inter-quartile range of Team X's score is (5.5 - 0) = 5.5.

<u>Now, we will calculate the mean, median, range and inter-quartile range for Team Y;</u>

Mean of Team Y data is given by the following formula;

Mean, $\bar Y$ = $\frac{\sum Y}{n}$

= $\frac{4+ 2+ 0+ 3+ 2+ 1+ 6+ 4}{8}$ = $\frac{22}{8}$ = 2.75

So, the mean of Team Y's scores is 2.75.

Now, for calculating the median; we have to arrange the data in ascending order and then observe that the number of observations (n) in the data is even or odd.

Team Y: 0, 1, 2, 2, 3, 4, 4, 6

If n is odd, then the formula for calculating median is given by;

Median = $(\frac{n+1}{2} )^{th} \text{ obs.}$

If n is even, then the formula for calculating median is given by;

Median = $\frac{(\frac{n}{2})^{th} \text{ obs.} +(\frac{n}{2}+1)^{th} \text{ obs.} }{2}$

Here, the number of observations is even, i.e. n = 8.

So, Median = $\frac{(\frac{n}{2})^{th} \text{ obs.} +(\frac{n}{2}+1)^{th} \text{ obs.} }{2}$

= $\frac{(\frac{8}{2})^{th} \text{ obs.} +(\frac{8}{2}+1)^{th} \text{ obs.} }{2}$

= $\frac{(4)^{th} \text{ obs.} +(5)^{th} \text{ obs.} }{2}$

= $\frac{2+3}{2}$ = 2.5

So, the median of Team Y's score is 2.5.

Now, the range is calculated as the difference between the highest and the lowest value in our data.

Range = Highest value - Lowest value

= 6 - 0 = 6

So, the range of Team Y's score is 6.

Now, the inter-quartile range of the data is given by;

Inter-quartile range = $Q_3-Q_1$

$Q_1=(\frac{n+1}{4} )^{th} \text{ obs.}$

= $(\frac{8+1}{4} )^{th} \text{ obs.}$

= $(2.25 )^{th} \text{ obs.}$

$Q_1 = 2^{nd} \text{ obs.} + 0.25[ 3^{rd} \text{ obs.} -2^{nd} \text{ obs.} ]$

= 1 + 0.25[2 - 1] = 1.25

$Q_3=3(\frac{n+1}{4} )^{th} \text{ obs.}$

= $3(\frac{8+1}{4} )^{th} \text{ obs.}$

= $(6.75 )^{th} \text{ obs.}$

$Q_3 = 6^{th} \text{ obs.} + 0.75[ 7^{th} \text{ obs.} -6^{th} \text{ obs.} ]$

= 4 + 0.75[4 - 4] = 4

So, the inter-quartile range of Team Y's score is (4 - 1.25) = 2.75.

Hence, the correct statement is:

C. Team Y’s scores have a lower mean value.

4 0

3 years ago

The point (1,-1) is on the terminal side of angle theta in standard position what are the values of sine cuisine and tangent of

pickupchik [31]

$\bf \textit{terminal point }\implies 
\begin{array}{lclcll}
(&1&,&-1)\\
&\uparrow &&\uparrow \\
&x&&y
\end{array}
\\\\\\
sin(\theta)=\cfrac{y}{r}
\qquad 
% cosine
cos(\theta)=\cfrac{x}{r}
\qquad 
% tangent
tan(\theta)=\cfrac{y}{x}\\\\
-----------------------------\\\\
\textit{so, what's the radius "r"?}\implies r^2=x^2+y^2\implies r=\sqrt{x^2+y^2}$

3 0

3 years ago

Why does everyone expect people to awnser questions but they never awnser any them selves like I need help and I've helped loads

Gnoma [55]

Answer:

thats crazy bro

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Bob is standing 25 feet from a lamppost that is to his left and 30 feet from a lamppost that is to his right. The distance betwe

irina1246 [14]

Answer:

The answer is 41

Step-by-step explanation:

400 = 900 + 625 - 1500 cos x

400 - 1525 = -1500 cos x

-1125 = -1500 cos

cos x = -1125/-1500

x = cos⁻¹ 0.75

x = 41.409622109

x ≈ 41°

3 0

2 years ago