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natima [27]
3 years ago
13

Bernardo graphed the amount of money he earned each week mowing lawns. A line graph titled Lawn Mowing Money has weeks on the x-

axis and money on the y-axis. Week 1 was 100 dollars, week 2 was 70 dollars, week 3 was 80 dollars, week 4 was 120 dollars, week 5 was 90 dollars, week 6 was 60 dollars. Between which weeks was the change in money earned the greatest? weeks 1 and 6 weeks 2 and 4 weeks 4 and 6 weeks 1 and 4
Mathematics
2 answers:
Sati [7]3 years ago
6 0

Answer:

My answer was deleted but it was between 4 and 6 weeks

Step-by-step explanation:

OverLord2011 [107]3 years ago
4 0

Answer:

4 and 6

Step-by-step explanation:

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A laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighings. Scale readings in repeated we
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Answer:

99% confidence interval for the given specimen is [3.4125 , 3.4155].

Step-by-step explanation:

We are given that a laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighing. Scale readings in repeated weighing are Normally distributed with mean equal to the true weight of the specimen.

Three weighing of a specimen on this scale give 3.412, 3.416, and 3.414 g.

Firstly, the pivotal quantity for 99% confidence interval for the true mean specimen is given by;

        P.Q. = \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \bar X = sample mean weighing of specimen = \frac{3.412+3.416+3.414}{3} = 3.414 g

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            n = sample of specimen = 3

            \mu = population mean

<em>Here for constructing 99% confidence interval we have used z statistics because we know about population standard deviation (sigma).</em>

So, 99% confidence interval for the population​ mean, \mu is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99  {As the critical value of z at 0.5% level

                                                            of significance are -2.5758 & 2.5758}

P(-2.5758 < \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } < 2.5758) = 0.99

P( -2.5758 \times {\frac{\sigma}{\sqrt{n} } } < {\bar X - \mu} < 2.5758 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.99

P( \bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } } < \mu < \bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.99

<u>99% confidence interval for</u> \mu = [ \bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } } , \bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } } ]

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Therefore, 99% confidence interval for this specimen is [3.4125 , 3.4155].

6 0
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