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Delicious77 [7]
2 years ago
14

In a sample of 100 steel canisters, the mean wall thickness was 8.1 mm with a standard deviation of 0.5 mm. Someone says that th

e mean thickness is less than 8.2 mm. With what level of confidence can this statement be made? (Express the final answer as a percent and round to two decimal places.)
Mathematics
1 answer:
Travka [436]2 years ago
8 0

Answer:

This statement can be made with a level of confidence of 97.72%.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 8.1 mm

Standard Deviation, σ = 0.5 mm

Sample size, n = 100

We are given that the distribution of thickness is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

Standard error due to sampling:

=\dfrac{\sigma}{\sqrt{n}} = \dfrac{0.5}{\sqrt{100}} = 0.05

P(mean thickness is less than 8.2 mm)

P(x < 8.2)

P( x < 8.2)\\\\ = P( z < \displaystyle\frac{8.2 - 8.1}{0.05})\\\\ = P(z < 2)

Calculation the value from standard normal z table, we have,  

P(x < 8.2) =0.9772 = 97.72\%

This statement can be made with a level of confidence of 97.72%.

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<u>Complete Question: </u>

The table (in figure) represents the recommended exercise intensity for the aerobic program based on the number of visits since the beginning program recommends a consistent intensity for three visits increase in intensity over three visits then decreasing intensity for three visits before restoring the pattern which set of values could be the intensity of embers exercise during the fourth fifth and six visits check all that apply

A.63%,63%,63%

B.66%,69%,72%

C.66%,62%,58%

D67%,67%,67%

E.63%,65%,67%

F.67%,72%,77%

<u>Answer: </u>

<u>The following set of values could be the intensity of embers exercise during the fourth fifth and six visits :</u>

  • 66%, 69%, 72%
  • 63%, 65%, 67%
  • 67%, 72%, 77%

<u> Step-by-step explanation:</u>

Given that intensity for first three visits on Amber's exercise in consistent way at the program beginning. Then it increases for 3 visits and decreases over 3 visits. In question, they asked us to find the intensity of Amber’s exercise for fourth, fifth and sixth alone.

As per given statement, first three visits are consistent in intensity (constant for visit 1, 2, 3).Then increases in intensity of Amber’s exercise for visits 4, 5, 6 (suits option B, E, F). Then it gets decreased for further visits (7, 8, 9).

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1 year ago
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Selling price of app per user = $10 per month

Total no. of user in first month = 10

To tal no. of user added per month = 7

We have to find how much money is earned in one year

So the formula is n/2 * (2a +(n-1)d)

Here we have n = 12 which is total no. of months

a = 100 , which is money earned in one month

d = 70 , which is increase in money earned per month

So money earned in a year comes out to be 5820

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Answer:

a) From the chart crated with Microsoft Excel, we have that the correlation coefficient, r = √0.8581 ≈ 0.93 to the nearest hundredth

The steps used includes

1) Inputting the given data into the cells on a Microsoft Excel spread sheet

2) Highlighting and sorting the data in the cells in order of increasing Rainfall

3) Generating a dot plot using the sorted data from above

4) Adding the trend line, Square of the linear regression, and the trend line equation

5) Adding the axis labels

(b) The correlation coefficient states that there is a strong positive correlation between the monthly rainfall and and Umbrella sales

Step-by-step explanation:

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1 year ago
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LUCKY_DIMON [66]

Answer:

y2 = C1xe^(4x)

Step-by-step explanation:

Given that y1 = e^(4x) is a solution to the differential equation

y'' - 8y' + 16y = 0

We want to find the second solution y2 of the equation using the method of reduction of order.

Let

y2 = uy1

Because y2 is a solution to the differential equation, it satisfies

y2'' - 8y2' + 16y2 = 0

y2 = ue^(4x)

y2' = u'e^(4x) + 4ue^(4x)

y2'' = u''e^(4x) + 4u'e^(4x) + 4u'e^(4x) + 16ue^(4x)

= u''e^(4x) + 8u'e^(4x) + 16ue^(4x)

Using these,

y2'' - 8y2' + 16y2 =

[u''e^(4x) + 8u'e^(4x) + 16ue^(4x)] - 8[u'e^(4x) + 4ue^(4x)] + 16ue^(4x) = 0

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Let w = u', then w' = u''

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Integrating this, we have

w = C1

But w = u'

u' = C1

Integrating again, we have

u = C1x

But y2 = ue^(4x)

y2 = C1xe^(4x)

And this is the second solution

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2 years ago
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