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Alenkinab [10]
3 years ago
8

The scale of scores for an IQ test are approximately normal with mean 100 and standard deviation 15. The organization MENSA, whi

ch calls itself the “high IQ society”, requires a score of 130 or higher. What percent of adults are between IQ of 100 and 130 ?
A)68%
B)47.5%
C)95%
D)81.5%
Mathematics
1 answer:
Zina [86]3 years ago
5 0

Answer:

B) 47.5%

Step-by-step explanation:

Refer to the normal distribution chart attached. If 130 is 2 standard deviations higher than the mean (ignore the numbers beneath the percentages), then by the empirical rule, this means that 34%+13.5%=47.5% of adults are between IQs of 100 and 130. Therefore, option B is correct.

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Answer:

We conclude that:

\sqrt{\left(-81\right)x^2}=9ix

Step-by-step explanation:

Given the radical expression

\sqrt{\left(-81\right)x^2}

simplifying the expression

\sqrt{\left(-81\right)x^2}

Remove parentheses:  (-a) = -a

\sqrt{\left(-81\right)x^2}=\sqrt{-81x^2}

Apply radical rule:   \sqrt{-a}=\sqrt{-1}\sqrt{a},\:\quad \mathrm{\:assuming\:}a\ge 0

                 =\sqrt{-1}\sqrt{81x^2}

Apply imaginary number rule:  \sqrt{-1}=i

                 =i\sqrt{81x^2}

Apply radical rule:   \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b},\:\quad \mathrm{\:assuming\:}a\ge 0,\:b\ge 0

                  =\sqrt{81}i\sqrt{x^2}

                  =9i\sqrt{x^2}

Apply radical rule:  \sqrt[n]{a^n}=a,\:\quad \mathrm{\:assuming\:}a\ge 0

                  =9ix

Therefore, we conclude that:

\sqrt{\left(-81\right)x^2}=9ix

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