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yulyashka [42]
3 years ago
15

What is 1 2/3 divided by 1 5/6

Mathematics
2 answers:
Vilka [71]3 years ago
8 0
Ve thirds Conversion a mixed number 1 5/ 6 to a improper fraction: 1 5/6 = 1 5/ 6 = 1 · 6 + 5/ 6 = 6 + 5/ 6 = 11/ 6 To find a new numerator: a) Multiply the whole number 1 by the denominator 6. Whole number 1 equally 1 * 6/ 6 = 6/ 6 b) Add the answer from previous step 6 to the numerator 5. New numerator is 6 + 5 = 11 c) Write a previous answer (new numerator 11) over the denominator 6. One and five sixths is eleven sixths Divide: 5/ 3 : 11/ 6 = 5/ 3 · 6/ 11 = 5 · 6/ 3 · 11 = 30/ 33 = 3 · 10 / 3 · 11 = 10/ 11 Dividing two fractions is the same as multiplying the first fraction by the reciprocal value of the second fraction. The first sub-step is to find the reciprocal (reverse the numerator and denominator, reciprocal of 11/ 6 is 6/ 11 ) of the second fraction. Next, multiply the two numerators. Then, multiply the two denominators. In the following intermediate step, cancel by a common factor of 3 gives 10/ 11 . In other words - five thirds divided by eleven sixths = ten elevenths.
Lemur [1.5K]3 years ago
5 0

Answer:

10/11

Step-by-step explanation:

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Answer:

a) \frac{(8)(30.23)^2}{20.09} \leq \sigma^2 \leq \frac{(8)(30.23)^2}{1.65}

363.90 \leq \sigma^2 \leq 4430.80

Now we just take square root on both sides of the interval and we got:

19.08 \leq \sigma \leq 66.56

b) For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56

Step-by-step explanation:

423.6, 487.3, 453.2, 402.9, 483.0, 477.7, 442.3, 418.4, 459.0

Part a

The confidence interval for the population variance is given by the following formula:

\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}

On this case we need to find the sample standard deviation with the following formula:

s=sqrt{\frac{\sum_{i=1}^8 (x_i -\bar x)^2}{n-1}}
And in order to find the sample mean we just need to use this formula:
[tex]\bar x =\frac{\sum_{i=1}^n x_i}{n}

The sample deviation for this case is s=30.23

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

df=n-1=9-1=8

The Confidence interval is 0.98 or 98%, the value of \alpha=0.02 and \alpha/2 =0.01, and the critical values are:

\chi^2_{\alpha/2}=20.09

\chi^2_{1- \alpha/2}=1.65

And replacing into the formula for the interval we got:

\frac{(8)(30.23)^2}{20.09} \leq \sigma^2 \leq \frac{(8)(30.23)^2}{1.65}

363.90 \leq \sigma^2 \leq 4430.80

Now we just take square root on both sides of the interval and we got:

19.08 \leq \sigma \leq 66.56

Part b

For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56

4 0
3 years ago
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