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BaLLatris [955]

3 years ago

8

A city of Punjab has a 15 percent chance of wet weather on any given day. What is the probability that it will take a week for i

t three wet weather on 3 separate days? Also find its Standard Deviation

1 answer:

lilavasa [31]3 years ago

3 0

Answer:

so the probability will be = 0.062

Standard deviation = 0.8925

Explanation:

The probability of rain = 15% = 15/100= 0.15

and the probability of no rain=q = 1-p= 1-0.15= 0.85

The number of trials = 7

so the probability will be

7C3 * ( 0.15)^3 (0.85)^4= 35* 0.003375 * 0.52200 =0.06166= 0.062

Taking this as binomial as the p and q are constant and also the trials are independent .

For a binomial distribution

Standard deviation = npq= 0.15 *0.85 *7= 0.8925

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iren2701 [21]

Answer:

d

Explanation:

According to me answer is d but gas expand more than others

6 0

2 years ago

Compute the kinetic energy of a proton (mass 1.67×10−27kg ) using both the nonrelativistic and relativistic expressions for spee

JulsSmile [24]

Answer:

The non-relativistic kinetic energy of a proton is $6.76\times10^{-12}\ J$

The relativistic kinetic energy of a proton is $7.25\times10^{-12}\ m/s$

Explanation:

Given that,

Mass of proton $m=1.67\times10^{-27}\ kg$

Speed $v= 9.00\times10^{7}\ m/s$

We need to calculate the kinetic energy for non relativistic

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

Put the value into the formula

$K.E=\dfrac{1}{2}\times1.67\times10^{-27}\times(9.00\times10^{7})^2$

$K.E=6.76\times10^{-12}\ J$

We need to calculate the kinetic energy for relativistic

Using formula of kinetic energy

$K.E=mc^2(\sqrt{(\dfrac{1}{1-\dfrac{v^2}{c^2}})}-1)$

$K.E=1.67\times10^{-27}\times(3\times10^{8})^{2}\cdot\left(\sqrt{\frac{1}{1-\frac{\left(9.00\times10^{7}\right)^{2}}{(3\times10^{8})^{2}}}}-1\right)$

$K.E=7.25\times10^{-12}\ m/s$

Hence, The non-relativistic kinetic energy of a proton is $6.76\times10^{-12}\ J$

The relativistic kinetic energy of a proton is $7.25\times10^{-12}\ m/s$

7 0

3 years ago

si se deja caer un carrito desde el punto mas alto de ua psta de coches cuya altura es de 1.4m cual es la velocidad maxima que p

forsale [732]

Answer:

v = 5.24[m/s]

Explanation:

Este problema se puede resolver por medio del principio de la conservación de la energía, donde la energía potencial es igual a la energía cinética. Es decir a medida que el carrito desciende su energía potencial disminuye, pero su energía cinética aumenta.

$E_{kin}=E_{pot}$

Donde:

$E_{kin}=\frac{1}{2} *m*v^{2} \\\\E_{pot}=m*g*h$

Ahora reemplazando:

$\frac{1}{2} *m*v^{2}=m*g*h\\\\0.5*v^{2}=9.81*1.4\\v=\sqrt{\frac{9.81*1.4}{0.5} } \\\\v=5.24[m/s]$

6 0

3 years ago

Why do different stars have different life cycles

lana66690 [7]

The answer is shape is determined. A stars life cycle is determined in its size

3 0

2 years ago

A car in an amusement park ride rolls without friction around a track (Fig. P7.42). The hB car starts from rest at point A at a

MrRissso [65]

Answer:

$h>\dfrac{5}{2}R$

Explanation:

Given that

Height = h

Radius = R

From energy conservation

$KE_A+U_A=KE_B+U_B$

At point B

The minimum speed to complete the the circle

$V_B=\sqrt{gR}\ m/s$

So the kinetic energy at point B

$KE_B=\dfrac{1}{2}mV^2$

$KE_B=\dfrac{1}{2}mgR$

$KE_A+U_A=KE_B+U_B$

$0+mgh=\dfrac{1}{2}mgR+2mgR$

Without falling off at the top (point B)

$0+mgh>\dfrac{1}{2}mgR+2mgR$

$mg(h-2R)>\dfrac{1}{2}mgR$

$g(h-2R)>\dfrac{1}{2}gR$

$h>\dfrac{5}{2}R$

6 0

3 years ago