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ValentinkaMS [17]

3 years ago

11

How does the motion of an airplane compare to the motion of a rocket ?

1 answer:

Dovator [93]3 years ago

5 0

The airplane engine uses oxygen from the air to keep the engines working. Rockets must carry their oxygen for the engines to operate. Third Law of Motion). Rocket propellants by combustion provide the hot gas es that produce reaction force (thrust) when ejected to the rear.

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Help ASAP plssssssssss

garik1379 [7]

C. 32400 because 540 * 60 = 32400

5 0

2 years ago

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

$p(xg)$

$p(x$

$p(x$

$p(x$

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

2 years ago

An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) whi

Veronika [31]

Answer:

$s=5.79\ km$

$\theta=47^{\circ}$ east of south

Explanation:

Given:

distance of the person form the initial position, $d'=8.4\ km$

direction of the person from the initial position, $47^{\circ}$ north of east

distance supposed to travel form the initial position, $d=5.3\ km$

direction supposed to travel from the initial position, due North

<u>Now refer the schematic for visualization of situation:</u>

$y=d'.\sin47^{\circ}-d$

$y=8.4\times \sin47-5.3$ ...............(1)

$x=d'.\cos47^{\circ}$

$x=8.4\times \cos47^{\circ}$ .................(2)

<u>Now the direction of the desired position with respect to south:</u>

$\tan\theta=\frac{y}{x}$

$\tan\theta=\frac{8.4\times \sin47}{8.4\times \cos47}$

$\theta=47^{\circ}$ east of south

<u>Now the distance from the current position to the desired position:</u>

$s=\sqrt{x^2+y^2}$

$s=\sqrt{(8.4\times \cos47)^2+(8.4\times \sin47-5.3)^2}$

$s=5.79\ km$

4 0

2 years ago

Does groundwater flow or stay still?

zhenek [66]

Answer:

yes it flows through flow paths.

Explanation:

5 0

2 years ago

15) A 1720 kg car accelerates at a rate of 3.0 m/s2. How much force is the car's engine producing? (use the formula F=ma

Anastasy [175]

Answer:

<h3>The answer is 5160 N</h3>

Explanation:

To find the force acting on an object given it's mass and acceleration we use the formula

<h3>Force = mass × acceleration</h3>

From the question

mass = 1720 kg

acceleration = 3.0 m/s²

We have

Force = 1720 × 3

We have the final answer as

<h3>5160 N</h3>

Hope this helps you

4 0

2 years ago

Read 2 more answers