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notsponge [240]
3 years ago
6

2m-1-5m+4 what is the answer?

Mathematics
1 answer:
KiRa [710]3 years ago
8 0
2m-1-5m+4 

(2m-5m)-1+4

 
   Gather \ like \ terms 

-3m-1+4 

-3m+3

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An automobile company wants to determine the average amount of time it takes a machine to assemble a car. A sample of 40 times y
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Answer:

A 98% confidence interval for the mean assembly time is [21.34, 26.49] .

Step-by-step explanation:

We are given that a sample of 40 times yielded an average time of 23.92 minutes, with a sample standard deviation of 6.72 minutes.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                               P.Q. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample average time = 23.92 minutes

             s = sample standard deviation = 6.72 minutes

             n = sample of times = 40

             \mu = population mean assembly time

<em> Here for constructing a 98% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation. </em>

<u>So, a 98% confidence interval for the population mean, </u>\mu<u> is; </u>

P(-2.426 < t_3_9 < 2.426) = 0.98  {As the critical value of z at 1%  level

                                               of significance are -2.426 & 2.426}  

P(-2.426 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 2.426) = 0.98

P( -2.426 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 2.426 \times {\frac{s}{\sqrt{n} } } ) = 0.98

P( \bar X-2.426 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+2.426 \times {\frac{s}{\sqrt{n} } } ) = 0.98

<u>98% confidence interval for</u> \mu = [ \bar X-2.426 \times {\frac{s}{\sqrt{n} } } , \bar X+2.426 \times {\frac{s}{\sqrt{n} } } ]

                                     = [ 23.92-2.426 \times {\frac{6.72}{\sqrt{40} } } , 23.92+2.426 \times {\frac{6.72}{\sqrt{40} } } ]  

                                    = [21.34, 26.49]

Therefore, a 98% confidence interval for the mean assembly time is [21.34, 26.49] .

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