Ask question

Ask question

All categories

2 years ago

6

A large group of people get together. each one rolls a die 180 times, and counts the number of 1's. about what percentage of the

se people should get counts in the range of 15-45?

1 answer:

Arte-miy333 [17]2 years ago

5 0

<span>Around 99% - 100%. This is because a die is six sided, which gives the odds of a one coming up being roughly 17% independent of every roll. 17% of 180 trials comes out to 30-31 times a one will show up every 180 trials. This puts you right in the middle of the 15-45 range which means that somebody will almost ALWAYS reach 15-45 one's in a trial of 180 rolls.</span>

You might be interested in

A 250 kg flatcar 25 m long is moving with a speed of 3.0 m/s along horizontal frictionless rails. A 61 kg worker starts walking

Anna11 [10]

Answer:

x=31.09m

Explanation:

p1=p2

The momentum of flatcar and the momentum of the worker so

The velocity of the worker is:

$m_{f}*v_{f}=m_{w}*v_{w}\\\\v_{f}=\frac{m_{f}*v_{f}}{m_{w}}\\v_{f}=\frac{61kg*3.0\frac{m}{s}}{250kg}\\v_{f}=0.732\frac{m}{s}$

The total motion has a total velocity and is

$Vt=v_{w}+v_{f}\\Vt=0.732\frac{m}{s}+3.0\frac{m}{s}\\Vt=3.732\frac{m}{s}$

The time the worker take walking is

$t=\frac{x}{v_{w}}\\t=\frac{25m}{3\frac{m}{s}}=8.33s$

Now the total time and the total velocity determinate the motion of tha flatcar how far has moved

$x=t*Vt\\x=8.33s*3.732\frac{m}{s} \\x=31.09m$

5 0

2 years ago

A tennis ball (m=0.060 kg) is moving horizontally at 20 m/s toward a tennis player who hits it straight back at 26 m/s. What is

Rus_ich [418]

Answer:

0.36 kg-m/s

Explanation:

Given that,

Mass of a ball, m = 0.06 kg

Initial velocity of the ball, u = 20 m/s

Final velocity of the ball, v = 26 m/s

We need to find the change in momentum of the tennis ball. It is equal to the final momentum minus initial momentum

$\Delta p=m(v-u)\\\\=0.06\times (26-20)\\\\=0.36\ kg-m/s$

So, the change in momentum of the ball is 0.36 kg-m/s.

4 0

2 years ago

A cube that has a volume of 1.00 m³ contains N distinguishable particles.

grandymaker [24]

Answer:

1 P = 0.5

2 P = 0.3

3 P = 0.01

Explanation:

The probability formula is

$P =V^N$

Where P is the probability V is the volume while N is the number of distinguishing particles

So for N = 1 and $V = 0.500m^3$

$P = (0.500m^3)^1\\$

= 0.5

For N = 1 and $V = 0.300m^3$

$P = (0.300m^3)^1$

= 0.3

For N = 1 and $V = 0.0100m^3$

$P =(0.0100m^3)^1$

= 0.01

6 0

2 years ago

OK brainly if you want hard here here are sixty cups on a table. If one falls down, then how many remain

vivado [14]

Answer:

If one cup falls down then there will be 59 cups left.

5 0

3 years ago

Read 2 more answers

7. A mother pushes her 9.5 kg baby in her 5kg baby carriage over the grass with a force of 110N @ an angle

jasenka [17]

Weight of the carriage $=(m+M)g =142.1\ N$

Normal force $=Fsin(\theta) + W = 197.1\ N$

Frictional force $=\mu N=27.59\ N$

Acceleration $=4.66\ m\ s^{-2}$

Explanation:

We have to look into the FBD of the carriage.

Horizontal forces and Vertical forces separately.

To calculate Weight we know that both the mass of the baby and the carriage will be added.

So Weight(W) $=(m+M)\times g =(9.5+5)\ kg \times 9.8 =142.1\ Newton\ (N)$

To calculate normal force we have to look upon the vertical component of forces, as Normal force is acting vertically.We have weight which is a downward force along with $F_x$ , force of $110\ N$ acting vertically downward.Both are downward and Normal is upward so Normal force $=Summation\ of\ both\ forces$

Normal force (N) $= Fsin(\theta)+W=110sin(30) + 142.1 =197.1\ N$

Frictional force (f) $=\mu N=0.14\times 197.1 =27.59\ N$

To calculate acceleration we will use Newtons second law.

That is Force is product of mass and acceleration.

We can see in the diagram that $F_y=Horizontal$ and $F_x=Vertical$ component of forces.

So Fnet = Fy(Horizontal) - f(friction) $= m\times a$

Acceleration (a) = $\frac{Fcos(\theta)-\mu N}{mass(m)} =\frac{(95.26-27.59)}{14.5}= 4.66\ m\ s{^2 }$

So we have the weight of the carriage, normal force,frictional force and acceleration.

3 0

2 years ago