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STALIN [3.7K]
3 years ago
8

I am stuck and I hate maths so please help me

Mathematics
1 answer:
Scilla [17]3 years ago
3 0

Answer:

h = 21.2 cm

Step-by-step explanation:

You know the measure of one acute angle. It is 23 deg.

For that acute angle, the 9-cm leg is the opposite leg, and the h-cm leg is the adjacent leg. The trig ratio that relates the opposite leg to the adjacent leg is the tangent.

\tan A = \dfrac{opp}{adj}

\tan 23^\circ = \dfrac{9}{h}

h \tan 23^\circ = 9

h = \dfrac{9}{\tan 23^\circ}

h = \dfrac{9}{0.4245}

h = 21.2

Answer: h = 21.2 cm

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A)Mean :10.65

Median =10.7

Mode : 10.7

B)Range = 3.5

Standard deviation :0.89916

C)The response time of 8.3 minutes should be considered an outlier in comparison to the other response times

Step-by-step explanation:

A)

Data : 11.8 ,10.3, 10.7, 10.6, 11.5, 8.3, 10.5, 10.9, 10.7, 11.2

Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\Mean = \frac{11.8 +10.3+ 10.7+ 10.6+ 11.5+ 8.3+ 10.5+ 10.9+ 10.7+ 11.2}{10}

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Median: The mid value of the data

Data in ascending order

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Median = \frac{(\frac{n}{2})\text{th term}+(\frac{n}{2}+1)\text{th term}}{2}\\Median = \frac{(\frac{10}{2})\text{th term}+(\frac{10}{2}+1)\text{th term}}{2}\\Median = \frac{10.7+10.7}{2}\\Median = 10.7

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B) Range = Maximum - Minimum=11.8-8.3=3.5

Standard deviation : \sqrt{\frac{\sum(x-\bar{x})^2}{n}}

Standard deviation :\sqrt{\frac{(11.8-10.65)^2+(10.3-10.65)^2+......+(10.9-10.65)^2+(10.7-10.65)^2+(11.2-10.65)^2}{10}}

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C)

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Range :(Q1-1.5IQR, Q3-1.5IQR)

Range :(10.5-1.5 \times 0.7, 11.2-1.5 \times 0.7)

Range :(9.45, 10.15)

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So, the response time of 8.3 minutes should be considered an outlier in comparison to the other response times

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