What are the 3 major uses of water? Describe each in 3-4 sentence

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

Is it possible to have a net torque when all of the forces sum to zero? Explain.

nexus9112 [7]

Answer:

Yes it is possible

Explanation:

When two equal magnitude forces are acting on the rod in opposite direction

Then the net force on the system is always zero in that case

so we will have

$F - F = 0$

now for the system net torque due to these forces is given by

$\tau = F(r_1) + F(r_2)$

here we know that

$r_1, r_2$ = distance of the forces from reference about which torque is measured

so here we can say that net force is zero on the system while torque is not zero

in all such case object will rotate about a fixed position with change angular speed

3 0

3 years ago

(4A) The mass of Earth is 5.972 * 10^24 kg, and the radius of Earth is 6,371 km.

faltersainse [42]

Answer:

x₁ = 345100 km

Explanation:

The direction of the attraction forces between the earth and the object, and between the moon and the object, are in opposite direction and (along the straight line between the centers of earth and moon) and as gravity is always attractive, the net force will become zero when both forces are equal. According to this:

Let call "x₁" distance between center of the earth and the object, and

"x₂" the distance between center of the moon and the object, Mt mass of the earth, Ml mass of the moon, m₀ mass of the object

we can express:

F₁ ( force between earth and the object )

F₁ = K * Mt * m₀/ ( x₁)² K is a gravitational constant

F₂ (force between mn and the object)

F₂ = K * Ml * m₀ / (x₂)²

Then:

F₁ = F₂ K*Mt*m₀ / x₁² = K*Ml*m₀ /x₂²

Or simplifying the expression

Mt/ x₁² = Ml/ x₂²

We know that x₁ + x₂ = 384000 Km then

x₁ = 384000 - x₂

Mt/( 384000 - x₂)² = Ml / x₂²

Mt * x₂² = Ml *( 384000 - x₂)²

We need to solve for x₂

Mt * x₂² = Ml *[ ( 384000)² + x₂² - 768000*x₂]

By substitution:

5.972*10∧24*x₂² = 7.348*10∧22 * [ 1.47*10∧11 ] + 7.348*10∧22*x₂² -

7.348*10∧22*768000*x₂

Simplifying by 10∧22

5.972*10²*x₂² = 7.348* [ 1.47*10∧11 ] + 7.348*x₂²- 7.348*768000*x₂

Sorting out

5.972*10²*x₂²- 7.348*x₂² = 10.80*10∧11 - 56,43* 10∧5*x₂

(597,2 - 7,348 )* x₂² = 10.80*10∧11 - 56.43*10∧5*x₂

590x₂² + 56.43*10∧5*x₂ - 10.80*10∧11 = 0

Is a second degree equation

x₂ = -56.43*10∧5 ± √3184*10∧10 + 25488*10∧11 / 1160

x₂ ₁ = -56.43*10∧5 + √3184*10∧10 + 25488*10∧11 / 1160

x₂ ₁ = -56.43*10∧5 + √3184*10∧10 + 254880*10∧10 / 1160

x₂ ₁ = -56.43*10∧5 + 10∧5 [ √3184 + 254880 ] /1160

x₂ ₁ = -56.43*10∧5 + 508* 10∧5 / 1160

x₂ ₁ = 451.27*10∧5/1160

x₂ ₁ = 4512.7*10∧4 /1160

x₂ ₁ = 3.89*10∧4 km (distance between the moon and the object)

x₂ ₁ = 38900 km

x₂ = 38900 km

We dismiss the other solution because is negative and there is not a negative distance

Then the distance between the earth and the object is:

x₁ = 384000 - x₂

x₁ = 384000 - 38900

x₁ = 345100 km

5 0

3 years ago

It is friction that provides the force for a car to accelerate, so for high-performance cars the factor that limits acceleration

bearhunter [10]

Answer:

About 12 seconds

Explanation:

6 0

3 years ago

What is one benefit of lifelong physical activity?

kvasek [131]

The answer would be C

5 0

3 years ago