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djverab [1.8K]
2 years ago
14

A machine used to fill​ gallon-sized paint cans is regulated so that the amount of paint dispensed has a mean of 137 ounces and

a standard deviation of 0.30 ounces. You randomly select 45 cans and carefully measure the contents. The sample mean of the cans is 136.9 ounces. Does the machine need to be​ reset? Explain your reasoning.
Mathematics
2 answers:
ElenaW [278]2 years ago
5 0
 <span>Standard error of mean = Sigma / sqrt n 
= 0.30 / sqrt 45 
= 0.30 / 6.71 
= 0.045 
Xbar - Mu = 138.9 - 139 = - 0.1 
- 0.1 / 0.045 = - 2.22 

YES, because the sample mean lies beyond 2 Standard deviations on the left side of mu.

If this was the appropriate answer mark as the brainliest!
-procklown</span>
tensa zangetsu [6.8K]2 years ago
3 0

Answer:

Step-by-step explanation:

Let X be the amount of paint dispensed by a machine used to fill gallon=sized paint cans.

X is Normal(137, 0.30)

Sample size = 45

Sample std error = 0.30/sqrt 45 =0.0447

x bar = 136.9

Create hypotheses as:

H0: sample mean = 137

Ha: sample mean not equal to 137

(Two tailed test)

Test statistic = (136.9-137)/0.0447

=-2.237

p value = 0.0257

Since p >0.01, at 99% level we accept null hypothesis

Machine need not be reset.

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Dominik [7]

Answer:

\boxed{ \bold{ \sf{ \: 1. \:  \:  \:  \:  \: 198}}}

\boxed{ \bold{ \sf{2. \:  \:  \:  \:  \:  - 8}}}

\boxed{ \bold{ \sf{ 3. \:  \:  \:  \:  \: \frac{64}{343} }}}

Step-by-step explanation:

1. Given, u = 20 , x = 4 , y = 7 , z = 10

\sf{ \frac{u}{z}  + x {y}^{2} }

⇒\sf{ \frac{20}{10}  + 4 \times  {7}^{2} }

⇒\sf{ \frac{20}{10}  + 4 \times 49}

⇒\sf{ \frac{20}{10}  + 196}

⇒\sf{ \frac{20 + 196 \times 10}{10} }

⇒\sf{ \frac{20 + 1960}{10} }

⇒\sf{ \frac{1980}{10} }

⇒\sf{198}

2. \sf{4( - 2)}

Multiplying or dividing a positive integer by any negative integer gives a negative integer

= - 8

3. \sf{( \frac{4}{7} ) ^{3} }

⇒\sf{( \frac{ {4}^{3} }{ {7}^{3} } })

⇒\sf{ \frac{4 \times 4 \times 4 }{7 \times 7 \times 7}}

⇒\sf{ \frac{64}{343} }

Hope I helped!

Best regards! :D

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