Calculate the power rating of a home appliance (in kilowatts) that uses 8 amps of current when

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

A stuffed toy with a mass of 0.900 kilograms sits on the edge of a bed at a height of 0.830 if the toy falls off the bed what wi

Olenka [21]

Mechanical energy (ME) is the sum of potential energy (PE) and kinetic energy (KE). When the toy falls, energy is converted from PE to KE, but by conservation of energy, ME (and therefore PE+KE) will remain the same.

Therefore, ME at 0.500 m is the same as ME at 0.830 m (the starting point). It's easier to calculate ME at the starting point because its just PE we need to worry about (but if we wanted to we could calculate the instantaneous PE and KE at 0.500 m too and add them to get the same answer).

At the start:

ME = PE = mgh
ME = 0.900 (9.8) (0.830)
ME = 7.32 J

8 0

2 years ago

If we heated the ball up and kept the ring room temperature, would the ball be able to fit through the ring? (1 point) Why or wh

Fed [463]

Answer:

no:

Explanation:

it would grow and no longer be able to fit through the loop due to the hot air expanding.

8 0

2 years ago

The flywheel is rotating with an angular velocity ω0 = 2.37 rad/s at time t = 0 when a torque is applied to increase its angular

nika2105 [10]

Answer:

ω = 12.023 rad/s

α = 222.61 rad/s²

Explanation:

We are given;

ω0 = 2.37 rad/s, t = 0 sec

ω =?, t = 0.22 sec

α =?

θ = 57°

From formulas,

Tangential acceleration; a_t = rα

Normal acceleration; a_n = rω²

tan θ = a_t/a_n

Thus; tan θ = rα/rω² = α/ω²

tan θ = α/ω²

α = ω²tan θ

Now, α = dω/dt

So; dω/dt = ω²tan θ

Rearranging, we have;

dω/ω² = dt × tan θ

Integrating both sides, we have;

(ω, ω0)∫dω/ω² = (t, 0)∫dt × tan θ

This gives;

-1[(1/ω_o) - (1/ω)] = t(tan θ)

Thus;

ω = ω_o/(1 - (ω_o × t × tan θ))

While;

α = dω/dt = ((ω_o)²×tan θ)/(1 - (ω_o × t × tan θ))²

Thus, plugging in the relevant values;

ω = 2.37/(1 - (2.37 × 0.22 × tan 57))

ω = 12.023 rad/s

Also;

α = (2.37² × tan 57)/(1 - (2.37 × 0.22 × tan 57))²

α = 8.64926751525/0.03885408979 = 222.61 rad/s²

6 0

3 years ago

1. According to paragraph 3 in the text, MOST of the electromagnetic waves from

Fed [463]

Answer:

Most of the EM waves from the sun that reach Earth are infrared waves, visible light, and UV radiation.

Explanation:

I hope this helps! Have a good day!

5 0

3 years ago