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2 years ago

5

The diameters of apples from a certain farm follow the normal distribution with mean 4 inches and standard deviation 0.4 inch. A

pples can be size-sorted by being made to roll over mesh screens. First the apples are rolled over a screen with mesh size 3.5 inches. This separates out all the apples with diameters less than 3.5 inches. Second, the remaining apples are rolled over a screen with mesh size 4.3 inches. Find the proportion of apples with diameters less than 3.5 inches.

2 answers:

jeka942 years ago

8 0

Answer:

P = 10%

Explanation:

Let,

µ = mean of the population = 4 inches

σ = standard deviation = 0.4 inches

X= a random variable that gives apple’s diameter

P [X < 3.5] = P-value is the probability of apples which will be separated after passing through the mesh screen

After putting all the values in the formula, we will find the value of Z-score from the chart.

P [X < 3.5] = P [ $\frac{X-4}{0.4}$ < $\frac{3.5-4}{0.4}$ ]

P [Z < -1.25] = 0.09680

About 10% of the apples possess diameters less than 3.5 inches.

trasher [3.6K]2 years ago

7 0

Answer:

The probability for the proportion of apples with diameters less than 3.5 inches is 0.1056.

Explanation:

For a normal distribution, z score is calculated using this formula: (X-μ)/σx

Where Mean, μ = 4

Standard deviation, σ = 0.400

The probability for the proportion of apples with diameters less than 3.5 inches is expressed below:

Probability = P(X<3.5)

= P(Z<-1.25)

Probability = 0.1056

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Answer:

The impact occured at a distance of 2478.585 meters from the person.

Explanation:

(After some research on web, we conclude that problem is not incomplete) The element "Part A" may lead to the false idea that question is incomplete. Correct form is presented below:

<em>A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 6.4 seconds apart. How far away did the impact occur? (Sound speed in the air: 343 meters per second, sound speed in concrete: 3000 meters per second)</em>

Sound is a manifestation of mechanical waves, which needs a medium to propagate themselves. Depending on the material, sound will take more or less time to travel a given distance. From statement, we know this time difference between air and concrete ( $\Delta t$ ), in seconds:

$\Delta t = t_{A}-t_{C}$ (1)

Where:

$t_{C}$ - Time spent by the sound in concrete, in seconds.

$t_{A}$ - Time spent by the sound in the air, in seconds.

By suposing that sound travels the same distance and at constant speed in both materials, we have the following expression:

$\Delta t = \frac{x}{v_{A}}-\frac{x}{v_{C}}$

$\Delta t = x\cdot \left(\frac{1}{v_{A}}-\frac{1}{v_{C}} \right)$

$x = \frac{\Delta t}{\frac{1}{v_{A}}-\frac{1}{v_{C}} }$ (2)

Where:

$v_{C}$ - Speed of the sound in concrete, in meters per second.

$v_{A}$ - Speed of the sound in the air, in meters per second.

$x$ - Distance traveled by the sound, in meters.

If we know that $\Delta t = 6.4\,s$ , $v_{C} = 3000\,\frac{m}{s}$ and $v_{A} = 343\,\frac{m}{s}$ , then the distance travelled by the sound is:

$x = \frac{\Delta t}{\frac{1}{v_{A}}-\frac{1}{v_{C}} }$

$x = 2478.585\,m$

The impact occured at a distance of 2478.585 meters from the person.

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2 years ago

A constant force of 20. newtons applied to a box causes it to move at a constant speed of 4.0 meters per second. How much work i

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The answer is 3) 480 joules

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2 years ago

A point charge q1 = 3.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 10 m.

UkoKoshka [18]

Answer:

a) 1.6 mN b) -1.6 mN c) -1.6 mN d) 1.6 mN

Explanation:

The electrostatic force between 2 point charges, obeys the Coulomb's Law, that can be expressed as follows:

F₁₂ = k*q₁*q₂/(r₁₂)² (in magnitude)

The direction of the force, is along the line that joins the charges (along the x axis) and as q₁ and q₂ are of the same sign, aims away from both charges.

a) So, for the force on q₂, we have:

F₁₂ = 9*18*10⁻⁵ N = 1.6 mN (positive as it is aiming in the positive x direction)

b) The force on q1, according to Newton's 3rd Law, is just equal and opposite to the one on q2:

F₂₁ = (-9*18*10⁻⁵) N = -1.6 mN (towards the negative x direction, away from q1)

c) If q₂ were -6.0 μC, the force will be the same in magnitude, but as now both charges have different signs, they wil attract each other, so the direction of the forces will be exactly the opposite to the first case:

F₁₂ = -1.6 mN (going towards the origin, where q₁ is located)

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3 years ago

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Kaylis [27]

Answer:

Vi = 0.055 m³ = 55 L

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ΔQ = ΔU + W

where,

ΔQ = Heat absorbed by the system = 52.5 J

ΔU = Change in Internal Energy = -102.5 J (negative sign shows decrease in internal energy of the system)

W = Work Done in Expansion by the system = ?

Therefore,

52.5 J = - 102.5 J + W

W = 52.5 J + 102.5 J

W = 155 J

Now, the work done in a constant pressure condition is given by:

W = PΔV

W = P(Vf - Vi)

where,

P = Constant Pressure = (0.5 atm)(101325 Pa/1 atm) = 50662.5 Pa

Vf = Final Volume of System = (58 L)(0.001 m³/1 L) = 0.058 m³

Vi = Initial Volume of System = ?

Therefore,

155 J = (50662.5 Pa)(0.058 m³ - Vi)

Vi = 0.058 m³ - 155 J/50662.5 Pa

Vi = 0.058 m³ - 0.003 m³

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3 years ago

A student at a window on the second floor of a dorm sees her physics professor walking on the sidewalk beside the building. she

klasskru [66]

Refer to the diagram shown below.

In order for the balloon to strike the professor's head, th balloon should drop by 18 - 1.7 = 16.3 m in the time at the professor takes to walk 1 m.
The time for the professor to walk 1 m is
t = (1 m)/(0.45 m/s) = 2.2222 s

The initial vertical velocity of the balloon is zero.
The vertical drop of the balloon in 2.2222 s is
h = (1/2)*(9.8 m/s²)*(2.2222 s)² = 24.197 m

Because 24.97 > 16.3, the balloon lands in front of the professor, and does not hit the professor.

The time for the balloon to hit the ground is
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The time difference is 2.2222 - 1.9166 = 0.3056 s
Within this time interval, the professor travels 0.45*0.3056 = 0.175 m
Therefore the balloon falls 0.175 m in front of the professor.

Answer:
The balloon misses the professor, and falls 0.175 m in front of the professor.

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3 years ago