How are the electric field lines around a positive charge affected when a second positive charge is near it?

In a large restaurant an average 60% customers ask for water with their meal. A random sample of 10 customers is selected. Find

Gekata [30.6K]

Answer:

a) $P(6)=0.25$

b) $p(x$

c) $p(x\geq3)=0.9878$

d) $\sigma=\sqrt{2.4}=1.5492$

Explanation:

From the question we are told that:

Population percentage $p_\%=\60%$

Sample size $n=10$

Let x =customers ask for water

Let y =customers dose not ask for water with their meal

Generally the equation for y is mathematically given by

$y=1-p_\%\\y=1-0.60\\y=0.40$

Generally the equation for pmf p(x) is mathematically given by

$P(x)=10C_x (0..6)^x(0.4)^{10-x}$

a)

Generally the probability that exactly 6 ask for water is mathematically given by

$P(x)6=10C_6 (0..6)^6(0.4)^{10-6}$

$P(6)=0.25$

b)

Generally the probability that less than 9 ask for water with meal is mathematically given by

$p(xg)$

$p(x$

c)

Generally the probability that at least 3 ask for water with meal is mathematically given by

$p(x\geq3)=1-p(x$

$p(x\geq3)=1-[p(0)+p(1)+p(2)]$

$p(x\geq3)=1-[0.00001+0.0015+0.0106]$

$p(x\geq3)=1-[0.0122]$

$p(x\geq3)=0.9878$

d)

Generally the mean and standard deviation of sample size is mathematically given by

Mean

$\=x=np=10(0.6)=6$

Standard deviation

$v(x)=npq=10(0.6)(0.4)=2.4$

$\sigma=\sqrt{2.4}=1.5492$

4 0

3 years ago

Suppose you design a new thermometer called the "x" thermometer. on the x scale, the boiling point of water is 130.0 ox and the

Hoochie [10]

You've told us:

-- 130°x = 212°F

and

-- 10°x = 32°F

Thank you. Those are two points on a graph of °x vs °F . With those, we can figure out the equation of the graph, and easily convert ANY temperature on one scale to the equivalent temperature on the other scale.

-- If our graph is going to have °x on the horizontal axis and °F on the vertical axis, then the two points we know are (130, 212) and (10, 32) .

-- The slope of the line through these two points is

Slope = (32 - 212) / (10 - 130)

Slope = (-180) / (-120)

Slope = 1.5

So far, the equation of the graph is

F = 1.5 x + (F-intercept)

Plug one of the points into this equation. I'll use the second point (10, 32) just because the numbers are smaller:

32 = 1.5 (10) + F-intercept

32 = 15 + (F-intercept)

F-intercept = 17

So the equation of the conversion graph is

F = 1.5 x + 17

There you are ! Now you can plug ANY x temperature in there, and the F temperature jumps out at you.

The question is asking what temperature is the same on both scales. This seems tricky, but it's not too bad. Whatever that temperature is, since it's the same on both scales, you can take the conversion equation, and write the same variable in BOTH places.

We can write [ x = 1.5x + 17 ], solve it for x, and the solution will be the same temperature in F too.

or

We can write [ F = 1.5F + 17 ], solve it for F, and the solution will be the same temperature in x too.

F = 1.5F + 17

Subtract F from each side: 0.5F + 17 = 0

Subtract 17 from each side: 0.5F = -17

Multiply each side by 2 : F = -34

That should be the temperature that's the same number on both scales.

Let's check it out, using our handy-dandy conversion formula (the equation of our graph):

F = 1.5x + 17

Plug in -34 for x:

F = 1.5(-34) + 17

F = -51 + 17

It works ! -34 on either scale converts to -34 on the other one too. If the temperature ever gets down to -34, and you take both thermometers outside, they'll both read the same number.

6 0

3 years ago

Heat transfers energy from a hot object to a cold object. Both objects are isolated from their surroundings. The change in entro

aniked [119]

To develop this problem we will start from the definition of entropy as a function of total heat, temperature. This definition is mathematically described as

$S = \frac{Q}{T}$

Here,

Q = Total Heat

T = Temperature

The total change of entropy from a cold object to a hot object is given by the relationship,

$\Delta S = \frac{Q}{T_{cold}}-\frac{Q}{T_{hot}}$

From this relationship we can realize that the change in entropy by the second law of thermodynamics will be positive. Therefore the temperature in the hot body will be higher than that of the cold body, this implies that this term will be smaller than the first, and in other words it would imply that the magnitude of the entropy 'of the hot body' will always be less than the entropy 'cold body'

Change in entropy $\Delta S_{hot}$ is smaller than $\Delta S_{cold}$