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Monica [59]
3 years ago
13

maria rewrites a fraction less than 1 as a decimal the numerator is a whole number greater than 0 for which the denominator will

the fraction always convert to a repeating decimal? please help
Mathematics
2 answers:
Mumz [18]3 years ago
5 0

Answer:

Given:

Fraction is less than 1.

Numerator of fraction is whole Number greater than 0.

Decimal Expansion is repeating.

We know that when a fraction is less than 1 then Denominator of the fraction is greater than numerator of the fraction.

Also, we know that Decimal expansion is repeating only when denominator is not in form of 2^n5^m.

So, Required Fractions are like \frac{10}{12}\:,\:\frac{2}{9}.

I am Lyosha [343]3 years ago
4 0
The answer is 11, can you mark me brainliest if this helps?
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How do I simplify the n^-6*n^3 topic "Negative and Zero Exponents"
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3 years ago
A machine that fills beverage cans is supposed to put 12 ounces of beverage in each can. The standard deviation of the amount in
erica [24]

Answer:

\chi^2 =\frac{10-1}{0.0144} 0.0217 =13.54

The degrees of freedom are:

df=n-1=10-1=9

Now we can calculate the p value using the alternative hypothesis:

p_v =2*P(\chi^2 >13.54)=0.279

Since the p value is higher than the signficance level assumed of 0.05 we have enough evidence to FAIL to reject the null hypothesis and there is no evidence to conclude that the true deviation differs from 0.12 ounces

Step-by-step explanation:

Assuming the following data:"12.14 12.05 12.27 11.89 12.06

12.14 12.05 12.38 11.92 12.14"

We can calculate the sample deviation with this formula:

s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

n=10 represent the sample size

\alpha=0.05 represent the confidence level  

s^2 =0.0217 represent the sample variance

\sigma^2_0 =0.12^2= 0.0144 represent the value to verify

Null and alternative hypothesis

We want to determine whether the standard deviation differs from 0.12 ounce, so the system of hypothesis would be:

Null Hypothesis: \sigma^2 = 0.0144

Alternative hypothesis: \sigma^2 \neq 0.0144

The statistic can be calculated like this;

\chi^2 =\frac{n-1}{\sigma^2_0} s^2

\chi^2 =\frac{10-1}{0.0144} 0.0217 =13.54

The degrees of freedom are:

df=n-1=10-1=9

Now we can calculate the p value using the alternative hypothesis:

p_v =2*P(\chi^2 >13.54)=0.279

Since the p value is higher than the signficance level assumed of 0.05 we have enough evidence to FAIL to reject the null hypothesis and there is no evidence to conclude that the true deviation differs from 0.12 ounces

7 0
3 years ago
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