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3 years ago

0.66666667 as a fraction

Mathematics

2 answers:

Ierofanga [76]3 years ago

7 0

Answer:

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The correct answer is: " $\frac{2}{3}$ " .

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Note: Let's assume that this given decimal value is really:

"0.66666666666666666....." ;

(non-terminating; but repeating the singing digit, "7" ; yet rounded to:

" 0.66666667 " ;

then: " 0.666... " = $\frac{666}{999}$ ;

= (666 ÷ 111) / (999 ÷ 111) ;

= 6/9 ;

= (6÷3)/(9÷3) ;

= 2/3 ;

= " $\frac{2}{3}$ "

→ which is our answer.

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Or: " 0.6666... " = 666 ÷ 999 = ? ;

using calculator :

= 0.66666667 ;

→ which is the "rounded off" number of "0.6666..." ;

— (infinitely repeating) ;

→ So 666 ÷ 999 =

→ " $\frac{666}{999}$ " ;

= " $\frac{2}{3}$ " ;

→ Either use a "fraction calculator" ;

or simplify as aforementioned.

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Hope this answer is helpful to you!

Best wishes to you in your academic pursuits

— and within the "Brainly" community!

____________________________________________

erastova [34]3 years ago

5 0

Answer:

2/3

Step-by-step explanation:

The correct answer would be 2/3. Since .333333334 is 1/3 since if you divided 1 over 3 it would be .33334. And since 0.6666 is double that you would just add one more number to the numerator!

I hope this helps! Please mark as brainliest! Good luck! :D

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Answer:

The 95% confidence interval would be given by $0.281 \leq \mu_1 -\mu_2 \leq 0.529$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X_1 =9.2$ represent the sample mean 1

$\bar X_2 =8.8$ represent the sample mean 2

n1=27 represent the sample 1 size

n2=30 represent the sample 2 size

$\sigma_1 =0.3$ population standard deviation for sample 1

$\sigma_2 =0.1$ population standard deviation for sample 2

$\mu_1 -\mu_2$ parameter of interest.

Solution to the problem

The confidence interval for the difference of means is given by the following formula:

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2}\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}}$ (1)

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =9.2-8.8=0.4$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $t_{\alpha/2}=1.96$