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anyanavicka [17]
3 years ago
6

A marketing executive is investigating whether this year’s advertising campaign has resulted in greater mean sales compared with

last year’s mean sales. The executive collects a random sample of 100 customer orders from a large population of orders and calculates the sample mean and sample standard deviation.
Physics
1 answer:
djverab [1.8K]3 years ago
7 0

We make use of a one-sample t-test for a population mean.

One-sample t-test for a population mean

Option B

<h3> Sample mean and Sample standard deviation</h3>

A Sample Standard Deviation is the root-mean square of the  data  minus the sample mean,

The sample mean is is the mean of the randomly selected sample

Therefore, For a data or sample where we have no information on population standard deviation and here only one sample group is compared, we make use of a one-sample t-test for a population mean

A one-sample t-test for a population mean

More on Probability

brainly.com/question/795909

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If 5.4 J of work is needed to stretch a spring from 15 cm to 21 cm and another 9 J is needed to stretch it from 21 cm to 27 cm,
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the natural length of the spring is 9 cm

Explanation:

let the natural length of the spring = L

For each of the work done, we set up an integral equation;

5.4 = \int\limits^{21-l}_{15-l} {kx} \, dx \\\\5.4 = [\frac{1}{2}kx^2 ]^{21-l}_{15-l}\\\\5.4 = \frac{k}{2} [(21-l)^2 - (15-l)^2]\\\\k = \frac{2(5.4)}{(21-l)^2 - (15-l)^2}  \ \ \ -----(1)

The second equation of work done is set up as follows;

9 = \int\limits^{27-l}_{21-l} {kx} \, dx \\\\9 = [\frac{1}{2}kx^2 ]^{27-l}_{21-l}\\\\9 = \frac{k}{2} [(27-l)^2 - (21-l)^2] \\\\k = \frac{2(9)}{(27-l)^2 - (21-l)^2} \ \ \ -----(2)

solve equation (1) and equation (2) together;

\frac{2(9)}{(27-l)^2 - (21-l)^2} = \frac{2(5.4)}{(21-l)^2 - (15-l)^2}\\\\\frac{2(9)}{2(5.4)} = \frac{(27-l)^2 - (21-l)^2}{(21-l)^2 - (15-l)^2}\\\\\frac{9}{5.4} = \frac{(729 - 54l+ l^2) - (441-42l+ l^2)}{(441-42l+ l^2) - (225 -30l+ l^2)} \\\\\frac{9}{5.4 } = \frac{288-12l}{216-12l} \\\\\frac{9}{5.4 } =\frac{12}{12}  (\frac{24-l}{18 -l})\\\\\frac{9}{5.4 } = \frac{24-l}{18 -l}\\\\9(18-l) = 5.4(24-l)\\\\162-9l = 129.6-5.4l\\\\162-129.6 = 9l - 5.4 l\\\\32.4 = 3.6 l\\\\l = \frac{32.4}{3.6} \\\\

l = 9 \ cm

Therefore, the natural length of the spring is 9 cm

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