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Zanzabum
3 years ago
8

The life in hours of a battery is known to be approximately normally distributed, with standard deviation σ = 1.25 hours. A rand

om sample of 10 batteries has a mean life of x overBar equals 40.5 hours.
(a) Is there evidence to support the claim that battery life exceeds 40 hours? Use α = 0.020.
(b) What is the P-value for the test in part (a)?
(c) What is the β-error for the test in part (a) if the true mean life is 42 hours?
(d) What sample size would be required to ensure that β does not exceed 0.1 if the true mean life is 44 hours?
Round your answers to three significant digits.


(a) The battery life ( IS ??? or IS NOT ??? ) significantly different greater than 40 hours at α = 0.020.
(b)p-value = _____?_____ Round your answer to three decimal places (e.g. 98.765).
(c)β = _____?_____ Round your answer to five decimal places (e.g. 98.76543).
(d) ____?_____batteries

Mathematics
1 answer:
Papessa [141]3 years ago
6 0

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

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Answer:

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Step-by-step explanation:

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Suppose quantity s is a length and quantity t is a time. Suppose the quantities v and a are defined by v = ds/dt and a = dv/dt.
finlep [7]

Answer:

a) v = \frac{[L]}{[T]} = LT^{-1}

b) a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}

c) \int v dt = s(t) = [L]=L

d) \int a dt = v(t) = [L][T]^{-1}=LT^{-1}

e) \frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}

Step-by-step explanation:

Let define some notation:

[L]= represent longitude , [T] =represent time

And we have defined:

s(t) a position function

v = \frac{ds}{dt}

a= \frac{dv}{dt}

Part a

If we do the dimensional analysis for v we got:

v = \frac{[L]}{[T]} = LT^{-1}

Part b

For the acceleration we can use the result obtained from part a and we got:

a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}

Part c

From definition if we do the integral of the velocity respect to t we got the position:

\int v dt = s(t)

And the dimensional analysis for the position is:

\int v dt = s(t) = [L]=L

Part d

The integral for the acceleration respect to the time is the velocity:

\int a dt = v(t)

And the dimensional analysis for the position is:

\int a dt = v(t) = [L][T]^{-1}=LT^{-1}

Part e

If we take the derivate respect to the acceleration and we want to find the dimensional analysis for this case we got:

\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}

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PLS HELP AND ANSWER QUICK
Phoenix [80]

Answer:

m< K =<u>60</u>

m< L =<u>90</u>

m< KMI =<u>30</u>

m< LMA =<u>150</u>

Step-by-step explanation:

hope it helps you

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