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3 years ago

6

A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 420 gram setting. It

is believed that the machine is underfilling the bags. A 24 bag sample had a mean of 414 grams with a standard deviation of 15. Assume the population is normally distributed. A level of significance of 0.1 will be used. Find the P-value of the test statistic.

1 answer:

natta225 [31]3 years ago

8 0

Answer:

P-value of the test statistic is 2.499

Explanation:

given data:

size sample is 24

sample mean is 414 gm

standard deviation is 15

Null hypotheis is

$H_0: \mu = 420 gm$

$H_1 \mu< 420$

level of significance is 0.01

from test statics

$t = \frac{\hat x - \mu}{\frac{s}{\sqrt{n}}}$

degree od freedom is = n -1

df = 24 -1 = 13

$t = \frac{414 - 420}{\frac{15}{\sqrt{24}}}$

t = -1.959

from t table critical value of t at 0.1 significane level and 23 degree of freedom is 2.499

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3 years ago

A speed boat increases its speed uniformly from vi = 20.0 m/s to vf = 30.0 m/s in a distance of 2.00 x 10^2m. (a) Draw a coordin

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a) See graph in attachment

b) The suvat equation to use is $v_f^2 - v_i^2 = 2as$

c) The acceleration is $a=\frac{v_f^2-v_i^2}{2s}$

d) The acceleration is $1.25 m/s^2$

e) The time needed is 8 s

Explanation:

a)

For this part, find in attachment the diagram representing this situation.

Since we are not given any particular direction for the motion, we choose the x-direction as the direction of motion of the boat.

Then we have the following:

- The initial position of the boat is $x_i = 0$ , the origin

- The final position of the boat is $x_f = 200 m$

- The initial velocity of the boat is $v_i = 20.0 m/s$

- The final velocity of the boat is $v_f = 30.0 m/s$

Note that the arrow representing the final velocity is longer than that of the initial velocity, since the final velocity is larger.

b)

The motion of the speed boat is a uniformly accelerated motion (motion at constant acceleration), therefore we can use one of the suvat equations. In this particular problem, we know the following quantities:

$v_i = 20.0 m/s$ , the initial velocity

$v_f = 30.0 m/s$ , the final velocity

$s = x_f - x_i = 200 m$ , the displacement of the boat

Therefore, the equation that best can be use to find the acceleration is

$v_f^2 - v_i^2 = 2as$

where

$a$ is the acceleration

c)

Now we have to solve the equation

$v_f^2 - v_i^2 = 2as$

In order to find the acceleration.

This can be done by dividing both terms by $2s$ : this way, we find

$\frac{v_f^2-v_i^2}{2s}=\frac{2as}{2s}$

And so the acceleration is

$a=\frac{v_f^2-v_i^2}{2s}$

d)

Now we can use the equation found in part c) in order to find the acceleration.

We have the following data:

$v_i = 20.0 m/s$ , the initial velocity

$v_f = 30.0 m/s$ , the final velocity

$s = x_f - x_i = 200 m$ , the displacement of the boat

And substituting into the equation,

$a=\frac{30^2-20^2}{2(200)}=1.25 m/s^2$

e)

In order to find the time it takes the boat to travel the given distance, we can use the following suvat equation:

$v_f = v_i + at$

where:

$v_i$ is the initial velocity

$v_f$ is the final velocity

a is the acceleration

t is the time

Here we have:

$v_i = 20.0 m/s$

$v_f = 30.0 m/s$

$a=1.25 m/s^2$

Solving for t, we find:

$t=\frac{v_f-v_i}{a}=\frac{30-20}{1.25}=8 s$

Learn more about accelerated motion:

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3 years ago

An insulated rigid tank contains 3 kg of H2O in the form of a saturated mixture of liquid and vapor at a pressure of 150 kPa and

Andre45 [30]

Answer:

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Explanation:

from steam tables

At 150 kPa

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specific entropy values are

Sf = 1.4337 kJ/kg K

Sfg = 5.789 kJ/kg

initial specific volume is calculated as

$v_1 = vf + x(vg - vf)$

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$s_1 = Sf + x(Sfg)$

$= 1.4337 + 0.25 \times 5.7894 = 2.88 kJ/kg K$

FROM STEAM Table

at 200 kPa

specific volume

Vf = 0.001061 m^3/kg

vg = 0.88578 m^3/kg

specific entropy values are

Sf = 1.5302 kJ/kg K

Sfg = 5.5698 kJ/kg

constant volume so $v_1 - v_2 = 0.29064 m^3/kg$

$v_2 = v_1 = vf + x(vg - vf)$

$=0.29064 = x_2(0.88578 - 0.001061)$

$x_2 = 0.327$

$s_2 = 1.5302 + 0.32 \times 5.5968 = 3.36035 kJ/kg K$

Change in entropy $\Delta s = m(s_2 - s_1)$

=3( 3.36035 - 2.88) = 1.44 kJ/kg

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4 years ago