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katrin2010 [14]

2 years ago

10

The "Garbage Project" at the University of Arizona reports that the amount of paper discarded by households per week is normally

distributed with mean 9.4 lb and standard deviation 4.2 lb. What percentage of households throw out at least 9 lb of paper a week? (Round your answer to one decimal place.)

1 answer:

Lorico [155]2 years ago

7 0

Answer:

54%

Explanation:

We are given that

S.D=4.2 lb

Mean= $\mu=9.4 lb$

We have to find the percentage of household throw out at least 9 lb of paper a week.

Normal distribution formula :

$P(X\geq a)=P(z\geq \frac{a-\mu}{\sigma})$

We have a=9

$P(X\geq 9)=P(z\geq \frac{9-9.4}{4.2})=P(z\geq -0.1)$

$P(X\geq 9)=1-P(z$

$P(X\geq 9)=1-0.4602=0.5398\times 100=54$ %

Hence, the percentage of household throw out at least of paper a week=54%

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A two-resistor voltage divider employing a 2-k? and a 3-k? resistor is connected to a 5-V ground-referenced power supply to prov

vesna_86 [32]

Answer:

circuit sketched in first attached image.

Second attached image is for calculating the equivalent output resistance

Explanation:

For calculating the output voltage with regarding the first image.

$Vout = Vin \frac{R_{2}}{R_{2}+R_{1}}$

$Vout = 5 \frac{2000}{5000}[/[tex][tex]Vout = 5 \frac{2000}{5000}\\Vout = 5 \frac{2}{5} = 2 V$

For the calculus of the equivalent output resistance we apply thevenin, the voltage source is short and current sources are open circuit, resulting in the second image.

so.

$R_{out} = R_{2} || R_{1}\\R_{out} = 2000||3000 = \frac{2000*3000}{2000+3000} = 1200$

Taking into account the %5 tolerance, with the minimal bound for Voltage and resistance.

if the -5% is applied to both resistors the Voltage is still 5V because the quotient has 5% / 5% so it cancels. to be more logic it applies the 5% just to one resistor, the resistor in this case we choose 2k but the essential is to show that the resistors usually don't have the same value. applying to the 2k resistor we have:

$Vout = 5 \frac{1900}{4900}\\Vout = 5 \frac{19}{49} = 1.93 V$

$Vout = 5 \frac{2100}{5100}\\Vout = 5 \frac{21}{51} = 2.05 V$

$R_{out} = R_{2} || R_{1}\\R_{out} = 1900||2850= \frac{1900*2850}{1900+2850} = 1140$

$R_{out} = R_{2} || R_{1}\\R_{out} = 2100||3150 = \frac{2100*3150 }{2100+3150 } = 1260$

so.

$V_{out} = {1.93,2.05}V\\R_{1} = {1900,2100}\\R_{2} = {2850,3150}\\R_{out} = {1140,1260}$

4 0

2 years ago

A Porsche 944 Turbo has a rated engine power of 217hp . 30% of the power is lost in the drive train, and 70% reaches the wheels.

scZoUnD [109]

Explanation:

(a) It is given that two-third of weight is over the drive wheels. So, mathematically, w = $\frac{2}{3}mg$ .

Hence, maximum force is expressed as follows.

$F_{max} = \mu_{s} \times w$

$m \times a_{max} = \mu_{s} (\frac{2}{3} mg)$

Hence, the maximum acceleration is calculated as follows.

$a_{max} = \frac{2}{3} \mu_{s} \times g$

= $\frac{2}{3} \times 1.00 \times 9.8 m/s^{2}$

= 6.53 $m/s^{2}$

Hence, the maximum acceleration of the Porsche on a concrete surface where μs = 1 is 6.53 $m/s^{2}$ .

(b) Since, 30% of the power is lost in the drive train. So, the new power is 70% of $P_{max}$ .

That is, new power = $0.7 \times P_{max}$

Now, the expression for power in terms of force and velocity is as follows.

P = $F_{max} \nu$

$0.7 P_{max} = ma_{max} \nu$

Therefore, speed of the Porsche at maximum power output is as follows.

$\nu = 0.7 \times \frac{P_{max}}{ma_{max}}$

= $0.7 \times \frac{217 hp \times \frac{746 W}{1 hp}}{1500 kg \times 6.53 m/s^{2}}$

= 11.568 m/s

= 11.57 m/s

Therefore, speed of the Porsche at maximum power output is 11.57 m/s.

(c) The time taken will be calculated as follows.

time = $\frac{\text{velocity}}{\text{acceleration}}$

= $\frac{11.57 m/s}{6.53 m/s^{2}}$

= 1.77 s

Therefore, the Porsche takes 1.77 sec until it reaches the maximum power output.

6 0

2 years ago

Derive an expression for the total mechanical energy of the system as the monkey reaches the top of the motion, Etop, in terms o

ipn [44]

Answer:

U = 0.5 * k *(x + d - h_max)^2 + m*g*h_max

Explanation:

Given:

- The extension in spring @ equilibrium = x m

- The spring constant = k

- The amount of distance pulled down = d

- mass of the toy = m

Find:

- The total mechanical energy E_top at the top position h_max in terms of the available variables.

Solution:

- First we need to determine the types of Energy that are in play:

- The Elastic potential Energy E_p in a spring is given:

E_p: 0.5 * k * (ext)

- In our case when the toy at the top most position h_max will have a net extension ext, by summing displacement of spring:

ext = Equilibrium + distance pulled - h_max = (x + d - h_max)

Hence, the elastic potential energy will be:

E_p = 0.5 * k *(x + d - h_max)^2

- The gravitational potential energy E_g is given by:

E_g = m*g*h_max

Where, bottom most position is taken as reference (datum).

- The kinetic Energy E_k is given by:

E_k = 0.5*m*v_top^2

- Since we know that the maximum height is reached when velocity is zero

Hence, E_k = 0.5*m*0^2 = 0.

The total Energy of the system U is sum of all energies and play:

U = E_p + E_k + E_g

U = 0.5 * k *(x + d - h_max)^2 + m*g*h_max

8 0

2 years ago

In certain cases, using both the momentum principle and energy principle to analyze a system is useful, as they each can reveal

SpyIntel [72]

Answer:

A) F_g = 26284.48 N

B) v = 7404.18 m/s

C) E = 19.19 × 10^(10) J

Explanation:

We are given;

Mass of satellite; m = 3500 kg

Mass of the earth; M = 6 x 10²⁴ Kg

Earth circular orbit radius; R = 7.3 x 10⁶ m

A) Formula for the gravitational force is;

F_g = GmM/r²

Where G is gravitational constant = 6.67 × 10^(-11) N.m²/kg²

Plugging in the relevant values, we have;

F_g = (6.67 × 10^(-11) × 3500 × 6 x 10²⁴)/(7.3 x 10⁶)²

F_g = 26284.48 N

B) From the momentum principle, we have that the gravitational force is equal to the centripetal force.

Thus;

GmM/r² = mv²/r

Making v th subject, we have;

v = √(GM/r)

Plugging in the relevant values;

v = √(6.67 × 10^(-11) × 6 x 10²⁴)/(7.3 x 10⁶))

v = 7404.18 m/s

C) From the energy principle, the minimum amount of work is given by;

E = (GmM/r) - ½mv²

Plugging in the relevant values;

E = [(6.67 × 10^(-11) × 3500 × 6 × 10²⁴)/(7.3 x 10⁶)] - (½ × 3500 × 7404.18)

E = 19.19 × 10^(10) J

5 0

2 years ago

Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to t

yarga [219]

Answer

The rate at which the magnetic field is changing is $[\frac{dB}{dt} ] = 0.000467 T/s$

Explanation

From the question we are told that

The electric field strength is $E = 3.5mV/m = 3.5 *10^{-3} \ V/m$

The radius is $r = 1.5 \ m$

The rate of change of the magnetic field is mathematically represented as

$\frac{d \phi }{dt} = \int\limits^{} {E \cdot dl}$

Where $dl$ is change of a unit length

$\frac{d \phi}{dt} = A * \frac{dB}{dt}$

Where A is the area which is mathematically represented as

$A = \pi r^2$

So

$E \int\limits^{} { dl} = ( \pi r^2) (\frac{dB}{dt} )$

$E L = ( \pi r^2) (\frac{dB}{dt} )$

where L is the circumference of the circle which is mathematically represented as

$L = 2 \pi r$

So

$E (2 \pi r ) = (\pi r^2 ) [\frac{dB}{dt} ]$

$E = \frac{r}{2} [\frac{dB}{dt} ]$

$[\frac{dB}{dt} ] = \frac{E}{ \frac{r}{2} }$

substituting values

$[\frac{dB}{dt} ] = \frac{3.5 *10^{-3}}{ \frac{15}{2} }$

$[\frac{dB}{dt} ] = 0.000467 T/s$

8 0

2 years ago