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PolarNik [594]
3 years ago
12

The data set shows the weights of pears, in grams, in a crate of fruit. 143, 146, 158, 159, 162, 166, 169, 170, 170, 193, 197 Wh

ich data values are outliers? Select each correct answer.
143

146

193

197
Mathematics
2 answers:
DENIUS [597]3 years ago
8 0
197 and 193 because you have to make a box plot and whiskers 
SashulF [63]3 years ago
6 0

Answer:

193, 197 are outliers of data.

Step-by-step explanation:

Given :  The data set  143, 146, 158, 159, 162, 166, 169, 170, 170, 193, 197 .

To find :  Which data values are outliers?

Solution : We have given that  143, 146, 158, 159, 162, 166, 169, 170, 170, 193, 197 .

We need to find Outliers from given data :

Outliers : "Outliers" are values that "lie outside" the other values.

we can see from the data 193 and 197 are lie outside the others values.

Therefore , 193, 197 are outliers of data.

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We wish to give a 90% confidence interval for the mean value of a normally distributed random variable. We obtain a simple rando
coldgirl [10]

Answer: (9.27025,\ 11.12975)

Step-by-step explanation:

Given : Sample size : n= 9

Degree of freedom = df =n-1 =8

Sample mean : \overline{x}=10.2

sample standard deviation : s= 1.5

Significance level ; \alpha= 1-0.90=0.10

Since population standard deviation is not given , so we use t- test.

Using t-distribution table , we have

Critical value = t_{\alpha/2, df}=t_{0.05 , 8}=1.8595

Confidence interval for the population mean :

\overline{x}\pm t_{\alpha/2, df}\dfrac{s}{\sqrt{n}}

90% confidence interval for the mean value will be :

10.2\pm (1.8595)\dfrac{1.5}{\sqrt{9}}

10.2\pm (1.8595)\dfrac{1.5}{3}

10.2\pm (1.8595)(0.5)

10.2\pm (0.92975)

(10.2-0.92975,\ 10.2+0.92975)

(9.27025,\ 11.12975)

Hence, the 90% confidence interval for the mean value= (9.27025,\ 11.12975)

6 0
3 years ago
En el triángulo de vértices: A(8,3) ,B(-6,5) y C(-2,-6), determine:
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Alinara [238K]
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The difference of the square of a number and 4 is equal to 3 times that number. Find the negative solution.
leonid [27]

The negative solution is x = –1.

Solution:

Let the number be x.

Square of a number = x²

3 times a number = 3x

Difference of square of a number and 4 = 3 times that number

\Rightarrow x^2-4=3x

Subtract 3x from both sides of the equation, we get

\Rightarrow x^2-3x-4=0

–3x can be written as x – 4x.

\Rightarrow x^2+x-4x-4=0

\Rightarrow( x^2+x)+(-4x-4)=0

1st bracket have common term x and bracket have –4 common term.

\Rightarrow x( x+1)-4(x+1)=0

Take common term x + 1 outside.

(x+1)(x-4)=0

<em>By zero factor principle, if AB = 0 then A = 0 or B = 0.</em>

x + 1 = 0,   x – 4 = 0

x = –1,   x = 4

The negative solution is x = –1.

7 0
4 years ago
You lose 2 points every time you forget to write your name on a test. You have forgotten to write your name 6 times. What intege
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