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topjm [15]
3 years ago
12

Find the unit rate kenny reads 5/8 pages in 2/3 minutes

Mathematics
1 answer:
Cerrena [4.2K]3 years ago
4 0
.9375 pages / 1 minute

or 15/16 pages for every 16/16 minute.
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What 281, 480,100 in word form
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Two hundred eighty one million four hundred eighty thousand one hundred
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Solve for t in terms of q, r, and s.<br> r = sqt<br> t=
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t=r/(sq)

Step-by-step explanation:

r=sqt

=> r/(sq)=t

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Square root of 69x420
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To find the standard deviation of the diameter of wooden​ dowels, the manufacturer measures 19 randomly selected dowels and find
loris [4]

Answer:

Option D -  [0.12

Step-by-step explanation:

Given : The manufacturer measures 19 randomly selected dowels and finds the standard deviation of the sample to be s=0.16.

To find : The​ 95% confidence interval for the population standard deviation sigma?

Solution :

Number of sample n=19

The degree of freedom is Df=n-1=19-1=18

The standard deviation of the sample is s=0.16

Applying chi-square table to find critical value,

Upper critical value of \chi^2 is UC=\chi(\frac{0.05}{2},18) = 31.5264

Lower critical value of \chi^2 is  

LC=\chi(1-\frac{0.05}{2},18) = 8.2307

Lower limit of the 95% confidence interval for the population variance

L=\frac{(df)\times (s^2)}{UC}

L=\frac{18\times (0.16^2)}{31.5264}

L=\frac{18\times0.0256}{31.5264}

L=\frac{0.4608}{31.5264}

L=0.0146

Upper limit of the 95% confidence interval for the population variance

U=\frac{(df)\times(s^2)}{LC}

U=\frac{18\times (0.16^2)}{8.2307}

U=\frac{18\times0.0256}{8.2307}

U=\frac{0.4608}{8.2307}

U=0.0559

So, The 95% confidence interval for the population variance is [0.0146, 0.0560]

Now, The 95% confidence interval for the population standard deviation is

[\sqrt{0.0146}

[0.1208

or  [0.12

Therefore, Option D is correct.

The 95% confidence interval for the population standard deviation is  [0.12

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2 years ago
Which statement is true?
Karo-lina-s [1.5K]

Answer:

A parallelograms are rectangles

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2 years ago
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