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Lapatulllka [165]
2 years ago
12

Complete this section.

Mathematics
2 answers:
Phoenix [80]2 years ago
6 0

Answer:

6)       \frac{8}{3} =2\frac{2}{3}

7)        \frac{16}{15}=1\frac{1}{15}

8)        -\frac{23}{9} =-2\frac{5}{9}

9)       -\frac{17}{2} =-8\frac{1}{2}

10)       -\frac{55}{12} = -4\frac{7}{12}

Step-by-step explanation:

First of all we need to know what an improper fraction is and a mixed number. An improper fraction sometimes referred to as "Top heavy fraction" because a larger number is in the numerator than the denominator. A mixed fraction is a way of writing improper fractions with 3 number, a whole number and the a usual fraction. When converting an improper fraction to a mixed fraction, the denominator multiplied by the whole number plus the numerator should give the numerator on the improper fraction so,

6)       \frac{8}{3} =2\frac{2}{3}

7)        \frac{16}{15}=1\frac{1}{15}

8)        -\frac{23}{9} =-2\frac{5}{9}

9)       -\frac{17}{2} =-8\frac{1}{2}

10)       -\frac{55}{12} = -4\frac{7}{12}

choli [55]2 years ago
3 0

<em>look</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>picture</em>

<em>H</em><em>ope</em><em> </em><em>it</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>you</em>

<em>G</em><em>ood</em><em> </em><em>luck</em><em> </em><em>on</em><em> </em><em>your</em><em> </em><em>assignment</em>

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A student wanted to construct a 95% confidence interval for the mean age of students in her statistics class. She randomly selec
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Answer:

19.1-3.355\frac{1.5}{\sqrt{9}}=17.42    

19.1+3.355\frac{1.5}{\sqrt{9}}=20.78    

And the best option would be:

C. [17.42,20.78]

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".

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\bar X=19.1 represent the sample mean

\mu population mean (variable of interest)

s=1.5 represent the sample standard deviation

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Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=9-1=8

Since the Confidence is 0.99 or 99%, the value of \alpha=0.01 and \alpha/2 =0.005, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,8)".And we see that t_{\alpha/2}=

Now we have everything in order to replace into formula (1):

19.1-3.355\frac{1.5}{\sqrt{9}}=17.42    

19.1+3.355\frac{1.5}{\sqrt{9}}=20.78    

And the best option would be:

C. [17.42,20.78]

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