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2 years ago

9

To test the hypothesis that the population mean mu=3. 6, a sample size n=14 yields a sample mean 4. 007 and sample standard devi

ation 0. 607. Calculate the p-value and choose the correct conclusion

1 answer:

PolarNik [594]2 years ago

5 0

The P value for the given data set is 25127. For finding P value, we have to must find the Z value.

<h3>How to get the z scores?</h3>

If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.

The Z value is calculated as;

$Z = \dfrac{X - \mu}{\sigma})$

Z = (X - μ) / σ

Z = (4.007 - 3.6) / 0.607

Z = 0.67051

The P value for the given data set is 25127.

Learn more about z-score here:

brainly.com/question/21262765

#SPJ1

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A high-speed flywheel in a motor is spinning at 450 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and

alexira [117]

Answer:

A) $\omega_f=17.503\ rad.s^{-1}$

B) $t=55.6822\ s$

C) $\theta=1312\ rad$

Explanation:

Given:

mass of flywheel, $m=40\ kg$

diameter of flywheel, $d=0.72\ m$

rotational speed of flywheel, $N_i=450\ rpm \Rightarrow \omega_i=\frac{450\times 2\pi}{60} =15\pi\ rad.s^{-1}$

duration for which the power is off, $t_0=35\ s$

no. of revolutions made during the power is off, $\theta=180\times 2\pi=360\pi\ rad$

<u>Using equation of motion:</u>

$\theta=\omega_i.t+\frac{1}{2} \alpha.t^2$

$360\pi=15\pi\times 35+\frac{1}{2} \times \alpha\times35^2$

$\alpha=-0.8463\ rad.s^{-2}$

Negative sign denotes deceleration.

A)

Now using the equation:

$\omega_f=\omega_i+\alpha.t$

$\omega_f=15\pi-0.8463\times 35$

$\omega_f=17.503\ rad.s^{-1}$ is the angular velocity of the flywheel when the power comes back.

B)

Here:

$\omega_f=0\ rad.s^{-1}$

Now using the equation:

$\omega_f=\omega_i+\alpha.t$

$0=15\pi-0.8463\times t$

$t=55.6822\ s$ is the time after which the flywheel stops.

C)

Using the equation of motion:

$\theta=\omega_i.t+\frac{1}{2} \alpha.t^2$

$\theta=15\pi\times 55.68225-0.5\times 0.8463\times 55.68225^2$

$\theta=1312\ rad$ revolutions are made before stopping.

3 0

3 years ago

Define any two characteristics of sound

Oliga [24]

Answer:

five characteristics: Wavelength, Amplitude, Time-Period, Frequency and Velocity or Speed

5 0

3 years ago

A 900-kg car cruising at a constant speed of 60 km/h is to accelerate to 100 km/h in 4 s. The additional power needed to achieve

kiruha [24]

To solve this problem we will apply the concepts related to power as a function of the change of energy with respect to time. But we will consider the energy in the body equivalent to kinetic energy. The change in said energy will be the difference between the two velocity data given by half of the mass. We will first convert the given units into an international system like this

Initial Velocity,

$V_i = 60km/h (\frac{1000m}{1km})(\frac{1h}{3600s})$

$V_i = 16.6667m/s$

Final Velocity,

$V_f = 100km/h (\frac{1000m}{1km})(\frac{1h}{3600s})$

$V_f = 27.7778m/s$

Now Power is defined as the change of Energy over the time,

$P = \frac{E}{t}$

But Energy is equal to Kinetic Energy,

$P = \frac{\frac{1}{2} m\Delta v^2}{t}$

$P = \frac{\frac{1}{2} m(v_f^2-v_i^2)}{t}$

Replacing,

$P = \frac{\frac{1}{2} (900)(27.7778^2-16.6667^2)}{4}$

$P = 56kW$

Therefore the correct answer is A.

8 0

3 years ago

What average force is required to stop a 980 kg car in 9.0 s if the car is traveling at 91 km/h ?

kozerog [31]

Newton's second law of motion:
F=ma
$a = \frac{v}{t}$
You convert from km/h to m/s by dividing by 3.6:
$v = 91 \div 3.6$
$v = 25.3 \: \frac{m}{s}$
Then a is:
$a = \frac{25.3}{9}$
$a = 2.8 \frac{m}{s {}^{2} }$
Then:
F=(980)(2.8)=2744 N

5 0

3 years ago

A nonlinear spring is used to launch a toy car. The car is pushed against the spring, compressing the spring 2.5 cm. The force t

dybincka [34]

Answer:

1.56 J

Explanation:

given,

Spring compression, x = 2.5 cm

Force exerts by the spring,

F = - k x

k = 5000 N/m

Potential energy stored = ?

energy stored in the spring

$PE = \dfrac{1}{2}kx^2$

$PE = \dfrac{1}{2}\times 5000\times 0.025^2$

PE = 1.56 J

Hence, the potential energy stored in the car is equal to 1.56 J.

5 0

3 years ago