A bump where the continent hits the ocean floor

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

natta225 [31]

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

8 0

3 years ago

You performed an experiment in which you measured the amount of water leaking through different types of roofs. For one roof, yo

Ostrovityanka [42]

Given:
m = 13.2 oz, the mass of water measured

Note that
$1 \, oz = (1 \, oz)*( \frac{1}{16} \, \frac{lb}{oz} )*( \frac{1}{2.2} \, \frac{kg}{lb} ) = 0.0284 \, kg$

Answer: 0.0284 kg

6 0

3 years ago

Read 2 more answers

In a double-slit interference experiment, the slit separation is 2.41 μm, the light wavelength is 512 nm, and the separation bet

Elina [12.6K]

Answer:

Using equation 2dsinФ=n*λ

given d=2.41*10^-6m

λ=512*10^-12m

θ=52.64 degrees

7 0

4 years ago

Read 2 more answers

A mail carrier leaves the post ofﬁce and drives 2.00 km to the north. He then drives in a direction 60.0° south of east for 7.00

Oxana [17]

Answer:

12.97 km

Explanation:

In order to find the resultant displacement, we have to resolve each of the 3 displacements along the x and y direction.

Taking north as positive y direction and east as positive x-direction, we have:

- Displacement 1: 2.00 km to the north

So

$A_x = 0\\A_y = +2.00 km$

- Displacement 2: 60.0° south of east for 7.00 km

So

$B_x=(7.00)(cos (-45^{\circ}))=4.95 km\\B_y = (7.00)(sin (-45^{\circ}))=-4.95 km$

- Displacement 3: 9.50 km 35.0° north of east

So

$C_x=(9.50)(cos 35^{\circ})=7.78 km\\C_y=(9.50)(sin 35^{\circ})=5.45 km$

So the net displacement along the two directions is:

$R_x=A_x+B_x+C_x=0+4.95+7.78=12.73 km\\R_y=A_y+B_y+C_y=2.00+(-4.95)+5.45=2.50 km$

So, the distance between the initial and final position is equal to the magnitude of the net displacement:

$R=\sqrt{R_x^2+R_y^2}=\sqrt{12.73^2+2.50^2}=12.97 km$

7 0

3 years ago

A cave rescue team lifts an injured spelunker directly upward and out of a sinkhole by means of a motor-driven cable. The lift i

Leya [2.2K]

Answer:

a) 323.4J

b) 0J

c) -323.4J

Explanation:

a) W=Fd

F=ma

solve for acc. using kinematics

v^2=vo^2+2a(x)

8.41=2a(12)

4.205=a(12)

0.35=a

F=(77)(0.35)

F=26.95N

W=26.95*12...... W=323.4J

b) No acceleration, thus no force, thus no work!

c) W=Fd

F=ma

find acc. using kinematics: v^2=vo^2+2a(x)

0=(2.9^2)+2a(12)

0=8.41+2a(12)

-8.41=2a(12)

-4.205=a(12)

-0.35=a

F=(77)(-0.35)

F=-26.95N

W=(-26.95)(12)

W=-323.4J

Yes, work can be negative!

5 0

3 years ago