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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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KE= 0.5(m)(v^2)

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On a vacation flight, you look out the window of the jet and wonder about the forces exerted on the window. Suppose the air outs
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Answer:

The pressure difference is 25.8kPa while the force exerted on the window by air pressure is 3.02 kN

Explanation:

Using the Bernoulli's equation

P_{in}+\frac{1}{2}\rho v_{in}^2=P_{out}+\frac{1}{2}\rho v_{out}^2

Here

  • Pin-Pout is the pressure difference which is to be calculated.
  • ρ is the density of air whose value is 1.29 kg/m^3
  • vin is the velocity of air inside which is 0 m/s
  • vout is the velocity of air outside which is 200 m/s

Substituting values in the above equation yields

                                   \Delta P=P_{in}-P_{out}\\\Delta P=\frac{1}{2}\rho (v_{in}^2-v_{out}^2)\\\Delta P=\frac{1}{2}\times 1.29 ( 200^2-0)\\\Delta P=25.8 kPa

So the pressure difference is 25.8kPa.

As force is given by

                                          F=PA\\F=\Delta P A\\

Here

  • ΔP is the pressure difference calculated above
  • A is the area of the window given as

                                       A=L \times W\\A=0.26 \times 0.45\\A=0.117 m^2

Now force is

                                      F=\Delta P A\\F=25.8 \tiems 10^3 \times 0.117\\F=3.02 kN

So the force exerted on the window by air pressure is 3.02 kN

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C

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