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dem82 [27]
3 years ago
14

A triangle has a perimeter of 11.5 feet. The side lengths are x feet, 2x – 2 feet, and 3x + 1.5 feet. What is the value of x?​

Mathematics
2 answers:
Julli [10]3 years ago
7 0

Answer:

perimeter = h + l + b \\  = x + 2x - 2 + 3x + 1.5 = 11.5 \\ x \: terms \: togather \\ x + 2x + 3 x- 2 + 1.5 = 11.5 \\ 6x - .5 = 11.5 \\ 6x = 11.5 + .5 \\ 6x = 12 \\ x =  \frac{12}{6}  \\ x = 2 \\ thank \: you

Ronch [10]3 years ago
4 0

Answer:

  • x = 2 feet

Step-by-step explanation:

<u>We know that the perimeter is the sum of side lengths.</u>

  • P = a + b + c

<u>Substitute the values and solve for x:</u>

  • x + 2x - 2 + 3x + 1.5 = 11.5
  • 6x - 0.5 = 11.5
  • 6x = 12
  • x = 12/6
  • x = 2
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Answer:

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

8.3

10.3

10.5

10.6

10.7

10.7

10.9

11.2

11.5

11.8

n=10(even)

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

Mode : the most occurring frequency

10.7 is occurring twice while others are occurring once

So, Mode is 10.7

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}}

Standard deviation :0.89916

C)

8.3

,10.3

,10.5

,10.6

,10.7

,10.7

,10.9

,11.2

,11.5

,11.8

For Q1 ( Median of lower quartile )

8.3

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,10.5

,10.6

,10.7

Median = \frac{n+1}{2}\text{th term} =\frac{5+1}{2}=3 \text{rd term}=10.5

For Q3( Median of Upper quartile )

10.7

,10.9

,11.2

,11.5

,11.8

Median = \frac{n+1}{2}\text{th term} =\frac{5+1}{2}=3 \text{rd term}=11.2

IQR = Q3-Q1=11.2-10.5=0.7

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)

8.3 does not lie in this interval

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

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