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Lapatulllka [165]
2 years ago
11

63. 26 Which expressions are equal to 23 ? 123 126 26 . 33 23 33​

Mathematics
1 answer:
liubo4ka [24]2 years ago
3 0

Hey there!

(6^3 * 2^6) / 2^3

= (6 * 6 * 6 * 2 * 2 * 2 * 2 * 2 * 2) / 2 * 2 * 2

= (36 * 6 * 4 * 4 * 4) / 4 * 2

= (216 * 16 * 4) / 8
= 3,456 * 4 / 8

= 13,824 / 8

= 1,728


Looking for something that gives you the result of: 1,728


Option A.
12^3

= 12 * 12 * 12

= 144 * 12

= 1,728

Option A. is. possible answer


Option B.
6^3

= 6 * 6 * 6

= 36 * 6

= 216

216 ≠ 1,728

Option B. is incorrect


Option C.
12^6

= 12 * 12 * 12 * 12 * 12 * 12

= 144 * 144 * 144

= 20,736 * 144

= 2,985,984

2,985,984 ≠ 1,728

Option C. is also incorrect


Option D.
2^6 * 2^3

= 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2

= 4 * 4 * 4 * 4 * 2

= 16 * 16 * 2

= 256 * 2

= 512
512 ≠ 1,728

Option D. is also incorrect

Option E.
2^3 * 3^3

= 2 * 2 * 2 * 3 * 3 * 3

= 4 * 2 * 9 * 3

= 8 * 27

= 216

216 ≠ 1,728

Option E. is also incorrect.


Therefore, the answer should be:

Option A. 12^3



Good luck on your assignment & enjoy your day!



~Amphitrite1040:)

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GarryVolchara [31]

Answer:

\mu_p -\sigma_p = 0.74-0.0219=0.718

\mu_p +\sigma_p = 0.74+0.0219=0.762

68% of the rates are expected to be betwen 0.718 and 0.762

\mu_p -2*\sigma_p = 0.74-2*0.0219=0.696

\mu_p +2*\sigma_p = 0.74+2*0.0219=0.784

95% of the rates are expected to be betwen 0.696 and 0.784

\mu_p -3*\sigma_p = 0.74-3*0.0219=0.674

\mu_p +3*\sigma_p = 0.74+3*0.0219=0.806

99.7% of the rates are expected to be betwen 0.674 and 0.806

Step-by-step explanation:

Check for conditions

For this case in order to use the normal distribution for this case or the 68-95-99.7% rule we need to satisfy 3 conditions:

a) Independence : we assume that the random sample of 400 students each student is independent from the other.

b) 10% condition: We assume that the sample size on this case 400 is less than 10% of the real population size.

c) np= 400*0.74= 296>10

n(1-p) = 400*(1-0.74)=104>10

So then we have all the conditions satisfied.

Solution to the problem

For this case we know that the distribution for the population proportion is given by:

p \sim N(p, \sqrt{\frac{p(1-p)}{n}})

So then:

\mu_p = 0.74

\sigma_p =\sqrt{\frac{0.74(1-0.74)}{400}}=0.0219

The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).

\mu_p -\sigma_p = 0.74-0.0219=0.718

\mu_p +\sigma_p = 0.74+0.0219=0.762

68% of the rates are expected to be betwen 0.718 and 0.762

\mu_p -2*\sigma_p = 0.74-2*0.0219=0.696

\mu_p +2*\sigma_p = 0.74+2*0.0219=0.784

95% of the rates are expected to be betwen 0.696 and 0.784

\mu_p -3*\sigma_p = 0.74-3*0.0219=0.674

\mu_p +3*\sigma_p = 0.74+3*0.0219=0.806

99.7% of the rates are expected to be betwen 0.674 and 0.806

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