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Twenty-five wooden beams were ordered or a construction project. The sample mean and he sample standard deviation were measured

aksik [14]

Answer:

Correct option: B. 90%

Explanation:

The confidence interval is given by:

$CI = [\bar{x} - z\sigma_{\bar{x}} , \bar{x}+z\sigma_{\bar{x}} ]$

If $\bar{x}$ is 190, we can find the value of $z\sigma_{\bar{x}}$ :

$\bar{x} - z\sigma_{\bar{x}} = 188.29$

$190 - z\sigma_{\bar{x}} = 188.29$

$z\sigma_{\bar{x}} = 1.71$

Now we need to find the value of $\sigma_{\bar{x}}$ :

$\sigma_{\bar{x}} = s / \sqrt{n}$

$\sigma_{\bar{x}} = 5/ \sqrt{25}$

$\sigma_{\bar{x}} = 1$

So the value of z is 1.71.

Looking at the z-table, the z value that gives a z-score of 1.71 is 0.0436

This value will occur in both sides of the normal curve, so the confidence level is:

$CI = 1 - 2*0.0436 = 0.9128 = 91.28\%$

The nearest CI in the options is 90%, so the correct option is B.

4 0

3 years ago

How could the location of tests affect the performance of a catapult ?

Luba_88 [7]

Answer:

It could affect how far the projectile travels

Explanation:

Facing Uphill: Moves less far

Downhill: Moves further

3 0

3 years ago

Read 2 more answers

An alloy is evaluated for potential creep deformation in a short-term laboratory experiment. The creep rate (ϵ˙) is found to be

cupoosta [38]

Answer:

Activation energy for creep in this temperature range is Q = 252.2 kJ/mol

Explanation:

To calculate the creep rate at a particular temperature

creep rate, $\zeta_{\theta} = C \exp(\frac{-Q}{R \theta} )$

Creep rate at 800⁰C, $\zeta_{800} = C \exp(\frac{-Q}{R (800+273)} )$

$\zeta_{800} = C \exp(\frac{-Q}{1073R} )\\\zeta_{800} = 1 \% per hour =0.01\\$

$0.01 = C \exp(\frac{-Q}{1073R} )$ .........................(1)

Creep rate at 700⁰C

$\zeta_{700} = C \exp(\frac{-Q}{R (700+273)} )$

$\zeta_{800} = C \exp(\frac{-Q}{973R} )\\\zeta_{800} = 5.5 * 10^{-2} \% per hour =5.5 * 10^{-4}$

$5.5 * 10^{-4} = C \exp(\frac{-Q}{1073R} )$ .................(2)

Divide equation (1) by equation (2)

$\frac{0.01}{5.5 * 10^{-4} } = \exp[\frac{-Q}{1073R} -\frac{-Q}{973R} ]\\18.182= \exp[\frac{-Q}{1073R} +\frac{Q}{973R} ]\\R = 8.314\\18.182= \exp[\frac{-Q}{1073*8.314} +\frac{Q}{973*8.314} ]\\18.182= \exp[0.0000115 Q]\\$

Take the natural log of both sides

$ln 18.182= 0.0000115Q\\2.9004 = 0.0000115Q\\Q = 2.9004/0.0000115\\Q = 252211.49 J/mol\\Q = 252.2 kJ/mol$

3 0

3 years ago

Diffrent ways to make a header above a window

Oliga [24]

Answer: double click at the top of the page. Or you can also go to home file and click add heading.

Explanation:

5 0

3 years ago

Thin film deposition is a process where: a)-elemental, alloy, or compound thin films are deposited onto a bulk substrate! b)-Pho

marshall27 [118]

Answer:

(A) elemental, alloy, or compound thin films are deposited on to a bulk substrate

Explanation:

In film deposition there is process of depositing of material in form of thin films whose size varies between the nano meters to micrometers onto a surface. The material can be a single element a alloy or a compound.

This technology is very useful in semiconductor industries, in solar panels in CD drives etc

so from above discussion it is clear that option (a) will be the correct answer

8 0

3 years ago