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Veronika [31]
2 years ago
9

A normally distributed data set has a mean of 0 and a standard deviation of 2. Which of the following is closest to the percent

of values between -4.0 and 2.0?
please help will give brainlist!!!

Mathematics
1 answer:
frutty [35]2 years ago
4 0

A normally distributed data set has a mean of 0 and a standard deviation of 2. The closest to the percent of values between -4.0 and 2.0 would be 84%.

<h3>What is the empirical rule?</h3>

According to the empirical rule, also known as the 68-95-99.7 rule, the percentage of values that lie within an interval with 68%, 95%, and 99.7% of the values lies within one, two, or three standard deviations of the mean of the distribution.

P(\mu - \sigma < X < \mu + \sigma)  \approx 68\%\\P(\mu - 2\sigma < X < \mu + 2\sigma)  \approx 95\%\\P(\mu - 3\sigma < X < \mu + 3\sigma)  \approx 99.7\%

A normally distributed data set has a mean of 0 and a standard deviation of 2.

Z=(x-\mu)/\sigma

P(x=-0.4)\\\\z=(-0.4-0)/2\\\\= -0.2\\\\P(x=2)\\z=(2-0)/2\\\\=1

P(-0.4 < x < 2)=p(-0.2 < z < 1)=p(-0.2 < z < 0)+p(0 < z < 1)……….(by symmetry)

=.49865+.3413

.83995…….(by (http://83995…….by) table value)

=.8400 × 100

=84%

Learn more about the empirical rule here:

brainly.com/question/13676793

#SPJ2

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gladu [14]
If a number is a multipule of 6 then it is a multiplule of 2 AND 3

if a number is a multiplule of 2 then the last digit should be divisible by 2 (ie, 0,2,4,6,8)

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try each

A. 333
last number is not divisible by 2 so not divisible by 6

B. 882
2 is divisible by 2
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18/3=6
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C. 424
4 is disibielbe by 2
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D. 106
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3 years ago
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Free_Kalibri [48]

Answer:

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Hence, ption B is true.

Step-by-step explanation:

Given the expression

\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}

solving the expression

\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}

as

\sqrt{x^2y^3}=xy\sqrt{y}

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so the expression becomes

\:\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}=xy\sqrt{y}+2xy^2\sqrt{x}+xy\sqrt{y}

Group like terms

                                        =2xy^2\sqrt{x}+xy\sqrt{y}+xy\sqrt{y}

Add similar elements

                                        =2xy^2\sqrt{x}+2xy\sqrt{y}

Therefore, we conclude that:

\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}=2xy^2\sqrt{x}+2xy\sqrt{y}

Hence, option B is true.

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