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Ainat [17]
4 years ago
13

What is the written form of 30.55

Mathematics
1 answer:
kirza4 [7]4 years ago
4 0
Thirty and fifty-five hundredth.
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Which of the following best describes the set of numbers shown below?
Vsevolod [243]

Answer:

spongebob at the bottom of the sea

Step-by-step explanation:

In a plain, robust, conversational style, the author known as “Elena Ferrante” has captivated readers worldwide with her chronicle of a complicated friendship between two women.

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3 years ago
Find the 95% confidence intervalfor the variance and standard deviation for the time ittakes a customer to place a telephone ord
Aliun [14]

Answer:

8.637 \leq \sigma^2 \leq 28.93

2.939 \leq \sigma \leq 5.379

Step-by-step explanation:

Data given and notation

s represent the sample standard deviation

\bar x represent the sample mean

n=23 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 .

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

The sample standard deviation for this case was s = 3.8

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

df=n-1=23-1 =22

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 tabele to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.025,22)" "=CHISQ.INV(0.975,22)". so for this case the critical values are:

\chi^2_{\alpha/2}=36.78

\chi^2_{1- \alpha/2}=10.98

And replacing into the formula for the interval we got:

\frac{(22)(3.8)^2}{36.78} \leq \sigma^2 \leq \frac{(22)(3.8)^2}{10.98}

8.637 \leq \sigma^2 \leq 28.93

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

2.939 \leq \sigma \leq 5.379

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3 years ago
What is the solution of the following system?​
Greeley [361]

Answer:

Step-by-step explanation:

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4 years ago
How can you find the median when the number is not in the middle
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You add the two numbers and whatever you get you then divide it by 2
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If there are approximately 6 1/2 hours in the school day, how many minutes are in the school day?
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Multiply 6 by 60 and then add 30

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