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ololo11 [35]
3 years ago
5

I will give you BRAINLIEST for the correct answer

Mathematics
1 answer:
masya89 [10]3 years ago
5 0

Answer:

d

Step-by-step explanation:

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12.4. Measure of Dispersion<br>2076 Q.No. 15 Find the standard deviation of: 4, 6, 8, 10, 12.​
Alenkinab [10]

Answer:

Standard deviation of given data = 3.16227

Step-by-step explanation:

<u><em>Step(i)</em></u>:-

Given sample size 'n' = 5

Given data  4, 6,8,10,12

Mean = \frac{4+6+8+10+12}{5} = 8

Mean of the sample x⁻ = 8

Standard deviation of the sample

                  S.D = \sqrt{\frac{Sum(x-x^{-} )^{2} }{n-1}}

<u><em>Step(ii)</em></u>:-

Given data

x          :         4      6       8       10      12

x-x⁻      :      4 - 8   6-8   8-8    10-8    12-8

(x-x⁻)   :        -4      -2     0          2        4

(x-x⁻)²  :        16     4       0         4        16  

 

  S.D = \sqrt{\frac{Sum(x-x^{-} )^{2} }{n-1}}

  S.D = \sqrt{\frac{16+4+0+4+16}{4}}

 S.D = √10 = 3.16227

<u><em> Final answer</em></u>:-

The standard deviation = 3.16227

8 0
2 years ago
Which shows the expression below in simplified form? (2.2 × 1012) + (1.7 × 109)
AleksAgata [21]
<span>(2.2 × 1012) + (1.7 × 109) = 2411.7</span>
7 0
3 years ago
Read 2 more answers
Find the value of x and y that will make each quadrilateral a parallelogram.​
Tatiana [17]

Answer:

the opposite Angles sum up to give 180

Step-by-step explanation:

3y +3y = 180

y=30

if y = 30 the x = 55

are you clear?

7 0
3 years ago
Read 2 more answers
Verify that the segments are parallel
kozerog [31]

Answer:

We know that lines AB and CD are parallel because if you expand them, we know they will never touch.

Step-by-step explanation:

4 0
3 years ago
Average precipitation for the first 7 months of the year, the average precipitation in toledo, ohio, is 19.32 inches. if the ave
Colt1911 [192]
Part A:

The probability that a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \  \frac{a-\mu}{\sigma}\right)

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that <span>a randomly selected year will have precipitation greater than 18 inches for the first 7 months is given by:

P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\  \\ =1-P\left(z\ \textless \ \frac{18-19.32}{2.44}\right) \\  \\ =1-P(z\ \textless \ -0.5410) \\  \\ =1-0.29426=\bold{0.7057}



Part B:

</span>The probability that an n randomly selected samples of a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\frac{\sigma}{\sqrt{n}}}\right)

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that <span>5 randomly selected years will have precipitation greater than 18 inches for the first 7 months is given by:

</span>P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{\frac{2.44}{\sqrt{5}}}\right) \\ \\ =1-P(z\ \textless \ -1.210) \\ \\ =1-0.1132=\bold{0.8868}
7 0
3 years ago
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