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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:

Part a)

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Part b)

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Explanation:

When ball is dropped from height h = 4.0 m

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$v_f^2 - v_i^2 = 2 a d$

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Part a)

Average acceleration is given as

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$a = \frac{6.26 - (-8.86)}{12.0 \times 10^{-3}}$

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Part B)

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3 years ago

The origin of an x axis is placed at the center of a nonconducting solid sphere of radius R that carries a charge +qsphere distr

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Answer:

$q=49Q/64$

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$x =16R/15$

Explanation:

See attached figure.

$E_{Q}=$ E due to sphere

$E_{q}=$ E due to particule

$E_{total}=E_{Q}-E_{q}=0$ (1)

according to the law of gauss and superposition Law:

$E_{Q}=E_{1}+E_{2}=E_{2}$ ; electric field due to the small sphere with r1=R/4

$E_{Q}=kq_{2}/(r_{1}^{2})=$

$q_{2}=density*4/3*pi*r_{1}^{3}=Q/(4/3*pi*R^{3})*4/3*pi*r_{1}^{3}=Q*r_{1}^{3}/R^{3}$

then: $E_{Q}=kq_{2}/(r_{1}^{2})=k*Q*r_{1}^{3}/(R^{3}*r_{1}^{2}) = kQ/(4*R^{2})$ (2)

on the other hand, for the particule:

$E_{q}=kq/(r_{p}^{2})$

$r_{p}=2R-R/4=7R/4$ ⇒ $E_{q}=16kq/(49R^{2})$ (3)

We replace (2) y (3) in (1):

$E_{total}=E_{Q}-E_{q}=0=kQ/(4*R^{2}) - 49kq/(16R^{2})$

$q=49Q/64$

--------------------

if R<x<2R AND $E_{total}=E_{Q}-E_{q}=0$

$E_{total}=E_{Q}-E_{q}=0=kQ/(x^{2}) - kq/(2R-x^{2})$

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$Q(2R-x^{2})=49/64*x^{2}$

solving:

$x_{1} =16R/15$

$x_{2} =16R$

but: R<x<2R

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Given Information:

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mass of slender rod = m = 0.120 kg

mass of sphere welded to one end = m₁ = 0.0200 kg

mass of sphere welded to another end = m₂ = 0.0700 kg (typing error in the question it must be 0.0500 kg as given at the end of the question)

Required Information:

Linear speed of the 0.0500 kg sphere = v = ?

Answer:

Linear speed of the 0.0500 kg sphere = 1.55 m/s

Explanation:

The velocity of the sphere can by calculated using

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Where I is the moment of inertia of the whole setup ω is the speed and ΔKE is the change in kinetic energy

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I = (1/12)mL²

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L/2 means that the spheres are welded at both ends of slender rod whose length is L.

The overall moment of inertia becomes

I = (1/12)mL² + (m₁+m₂)(L/2)²

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I = 0.02227 kg.m²

The change in the potential energy is given by

ΔPE = m₁gh₁ + m₂gh₂

Where h₁ and h₂ are half of the length of slender rod

L/2 = 0.90/2 = 0.45 m

ΔPE = 0.0200*9.8*0.45 + 0.0500*9.8*-0.45

The negative sign is due to the fact that that m₂ is heavy and it would fall and the other sphere m₁ is lighter and it would will rise.

ΔPE = -0.1323 J

This potential energy is then converted into kinetic energy therefore,

ΔKE = ½Iω²

0.1323 = ½(0.02227)ω²

ω² = (2*0.1323)/0.02227

ω = √(2*0.1323)/0.02227

ω = 3.45 rad/s

The linear speed is

v = (L/2)ω

v = (0.90/2)*3.45

v = 1.55 m/s

Therefore, the linear speed of the 0.0500 kg sphere as its passes through its lowest point is 1.55 m/s.

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