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3 years ago

5

You weld two piece's of steel together. The toral thickness is 33/64 inch. If one piece of steel is 9/32 inch thick, how thick t

he other piece

1 answer:

Anon25 [30]3 years ago

4 0

Subtract the thickness of the known piece from the total thickness.

Because they have different denominators rewrite them to have the same.

9/32 x 2 = 18/64

Now you have 33/64 - 18/64 = 15/64

The other piece is 15/64 inch thick.

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What is -5n+7<57 and 6n+2<8

pishuonlain [190]

Answer:

if -5n+7<57, then n>-10.

if 6n+2<8, then n<1.

Step-by-step explanation:

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1. subtract 7 from both sides:

-5n+7 -7< 57 -7

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3 years ago

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 $[0.12$

6 0

3 years ago