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sukhopar [10]
3 years ago
10

Would someone be willing to help me? Please answer honestly and follow the directions, thank you!

Mathematics
1 answer:
larisa86 [58]3 years ago
8 0

Answer:

Scale factor is 10:1

Step-by-step explanation:

If you divide 1,450 by 145 you get 10. Then a scale factor is written in the form of a ratio. Therefor the scale factor is 10:1

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Read 2 more answers
The following observations were made on fracture toughness of a base plate of 18% nickel maraging steel (in ksi √in, given in in
Grace [21]

Answer:

A 90% confidence interval for the standard deviation of the fracture toughness distribution is [4.06, 6.82].

Step-by-step explanation:

We are given the following observations that were made on fracture toughness of a base plate of 18% nickel maraging steel below;

68.6, 71.9, 72.6, 73.1, 73.3, 73.5, 75.5, 75.7, 75.8, 76.1, 76.2,  76.2, 77.0, 77.9, 78.1, 79.6, 79.8, 79.9, 80.1, 82.2, 83.7, 93.4.

Firstly, the pivotal quantity for finding the confidence interval for the standard deviation is given by;

                             P.Q.  =  \frac{(n-1) \times s^{2} }{\sigma^{2} }  ~ \chi^{2} __n_-_1

where, s = sample standard deviation = \sqrt{\frac{\sum (X - \bar X^{2}) }{n-1} } = 5.063

            \sigma = population standard deviation

            n = sample of observations = 22

Here for constructing a 90% confidence interval we have used One-sample chi-square test statistics.

<u>So, 90% confidence interval for the population standard deviation, </u>\sigma<u> is ;</u>

P(11.59 < \chi^{2}__2_1 < 32.67) = 0.90  {As the critical value of chi at 21 degrees  

                                                  of freedom are 11.59 & 32.67}  

P(11.59 < \frac{(n-1) \times s^{2} }{\sigma^{2} } < 32.67) = 0.90

P( \frac{ 11.59}{(n-1) \times s^{2}} < \frac{1}{\sigma^{2} } < \frac{ 32.67}{(n-1) \times s^{2}} ) = 0.90

P( \frac{(n-1) \times s^{2} }{32.67 } < \sigma^{2} < \frac{(n-1) \times s^{2} }{11.59 } ) = 0.90

<u>90% confidence interval for</u> \sigma^{2} = [ \frac{(n-1) \times s^{2} }{32.67 } , \frac{(n-1) \times s^{2} }{11.59 } ]

                                     = [ \frac{21 \times 5.063^{2}  }{32.67 } , \frac{21 \times 5.063^{2}  }{11.59 } ]

                                     = [16.48 , 46.45]

<u>90% confidence interval for</u> \sigma = [\sqrt{16.48} , \sqrt{46.45} ]

                                                 = [4.06 , 6.82]

Therefore, a 90% confidence interval for the standard deviation of the fracture toughness distribution is [4.06, 6.82].

5 0
3 years ago
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