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NeX [460]
3 years ago
9

What the factor os 12​

Mathematics
1 answer:
Mariana [72]3 years ago
7 0

Answer:

Factors of 12 =1, 2, 3, 4, 6, 12

Step-by-step explanation:

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Set up and solve a quadratic equation to work out the value of x
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X= 24 simple calculations are there
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3 years ago
The quality control manager of a chemical company randomly sampled twenty 100-pound bags of fertilizer to estimate the variance
lisabon 2012 [21]

Answer:

The 95% confidence interval for the variance in the pounds of impurities would be 3.829 \leq \sigma^2 \leq 14.121.

Step-by-step explanation:

1) Data given and notation

s^2 =6.62 represent the sample variance

s=2.573 represent the sample standard deviation

\bar x represent the sample mean

n=20 the sample size

Confidence=95% or 0.95

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 mean or variance 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.

The Chi square distribution is the distribution of the sum of squared standard normal deviates .

2) Calculating the confidence interval

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

On this case the sample variance is given and for the sample deviation is just the square root of the sample variance.

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

df=n-1=20-1=19

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 values.  

The excel commands would be: "=CHISQ.INV(0.025,19)" "=CHISQ.INV(0.975,19)". so for this case the critical values are:

\chi^2_{\alpha/2}=32.852

\chi^2_{1- \alpha/2}=8.907

And replacing into the formula for the interval we got:

\frac{(19)(6.62)}{32.852} \leq \sigma^2 \frac{(19)(6.62)}{8.907}

3.829 \leq \sigma^2 \leq 14.121

So the 95% confidence interval for the variance in the pounds of impurities would be 3.829 \leq \sigma^2 \leq 14.121.

4 0
3 years ago
Explain how to determine which numbers must be excluded from the domain of a rational expression. Please provide an example with
Katarina [22]

Answer:

  • When we are having a rational expression i.e. a expression of the type:

\dfrac{f(x)}{g(x)}

Where f(x) and g(x) are polynomial functions.

Now the domain of this rational expression is whole of the real numbers except the points where the function g(x) will be zero.

Hence we have to exclude the points where the given denominator quantity is zero.

  • Let us consider an example as:

  Let R(x) denote the rational function as:

R(x)=\dfrac{x^2}{(x-2)(x-3)}

Now the domain of this rational function will be whole of the real line minus the points where the denominator is zero.

We know that (x-2)(x-3) is zero when x=2 or x=3.

Hence, the domain of R(x) is: R- {2,3}.

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3 years ago
What's Is the difference
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3 years ago
Plz help i wil mark u banlyest
Lapatulllka [165]

Answer:

B

Step-by-step explanation:

when you flip across the y axis you change the signs of the x coordinate but not the y coordinate.

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3 years ago
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