1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
ZanzabumX [31]
3 years ago
13

While conducting experiments, a marine biologist selects water depths from a uniformly distributed collection that vary between

2.00 m and 7.00 m. What is the probability that a randomly selected depth is between 2.25 m and 5.00 m?
Mathematics
1 answer:
aleksandr82 [10.1K]3 years ago
6 0

Answer:

The probability that a randomly selected depth is between 2.25 m and 5.00 m is 0.55.

Step-by-step explanation:

Let the random variable <em>X</em> denote the water depths.

As the variable water depths is continuous variable, the random variable <em>X</em> follows a continuous Uniform distribution with parameters <em>a</em> = 2.00 m and <em>b</em> = 7.00 m.

The probability density function of <em>X</em> is:

f_{X}(x)=\frac{1}{b-a};\ a

Compute the probability that a randomly selected depth is between 2.25 m and 5.00 m as follows:

P(2.25

                               =\frac{1}{5.00}\int\limits^{5.00}_{2.25} {1} \, dx\\\\=0.20\times [x]^{5.00}_{2.25} \\\\=0.20\times (5.00-2.25)\\\\=0.55

Thus, the probability that a randomly selected depth is between 2.25 m and 5.00 m is 0.55.

You might be interested in
Which is the answer
coldgirl [10]

Answer:b

Step-by-step explanation:

6 0
2 years ago
A uniform bar of mass $m$ and length $l$ is suspended on a frictionless hinge. A horizontally launched blob of clay of mass $m$
Irina-Kira [14]

Answer:

conservation of angular momentum ; conservation of energy

Step-by-step explanation:

The complete Question is given as follows:

" A uniform bar of mass (m) and length (L) is suspended on a frictionless hinge. A horizontally launched blob of clay of mass (m) strikes the bottom end of the bar and sticks to it. After that, the bar swings upward. What is the minimum initial speed (v) of the blob of clay that would enable the rod to swing a full circle? Which concepts/laws would be most helpful in solving this problem? Select the best answer from the options below.  "

CHOICES

kinematics of rotational motion; conservation of energy

conservation of momentum ; conservation of energy

conservation of angular momentum ; conservation of momentum

conservation of angular momentum ; conservation of energy

conservation of energy ; Newton's laws

Newton's laws ; conservation of angular momentum

conservation of angular momentum ; kinematics of rotational motion

Newton's laws, kinematics of rotational motion

Solution:

- We will apply the conservation of angular momentum M. Note the linear momentum does not remains conserved as the rod stores some energy as the clay sticks to the rod:

                                       M_i = M_f

- Initially the rod was at rest and clay had velocity of v, then M_i can be written as:

                                       M_i = m*v*L

- The final momentum is the combined effect of clay and rod:

                                       M_f = ( m*L^2 + I_rod )*w

- Where w is the angular speed of the rod after impact. And I_rod is the moment of inertia of rod.

                                      I_rod = mL^2 / 3

                                      M_f = ( m*L^2 + m*L^2 /3 )*w = (4*m*L^2 / 3)*w

- Formulate w in terms of initial velocity v:

                                     m*v*L = (4*m*L^2 / 3)*w

                                      0.75*v / L = w

- The minimum amount of velocity required would be enough to complete half of a circle.

- Apply conservation of Energy principle:

                                     T_i + V_i = T_f + V_f

Where, T is the kinetic energy soon after impact and at top most position.

            Assuming, T_f = 0 , for minimum velocity required to complete on circle.

             T_i = 0.5*I_combined*w^2

Where, I_combined = I_clay + I_rod = 4*m*L^2 / 3

And w = 0.75*v / L:

             T_i = 0.5*[4*m*L^2 / 3]*[0.75*v / L]^2

             T_i = 0.5*[m]*[v^2]

Also, V is the potential energy of the clay plus rod system soon after impact and at top most point.

             V_i = 0 ( Datum )

             V_f = V_rod + V_clay

             V_f = m*g*L + m*g*2L = 3*m*g*L

- Plug in the expressions in the energy balance and we get:

                              0.5*[m]*[v^2] + 0 = 0 + 3*m*g*L

                                      v_min = sqrt ( 6*g*L)

- So the choices used were:

conservation of angular momentum ; conservation of energy

                 

             

6 0
3 years ago
The function T(n) is defined below. Which of the following are equal to T(8)? Check all that apply.
Liula [17]
In function T(n), in which n = 8, plug in 8 for n

T(n) = 4n - 5
T(8) = 4(8) - 5

multiply 4 and 8

4(8) = 32

T(8) = 32 - 5

Simplify

T(8) = 27

27, or C) is your answer.


hope this helps
7 0
3 years ago
Read 2 more answers
Question 5
Semenov [28]

Answer:

NEITHER OF THEM IS THE ANSWER

6 0
3 years ago
CAN SOMEONE CHECK MY ANSWER
zalisa [80]

Answer:

Step-by-step explanation:

Simplify by combining similar terms: 2x+4-3x-6

8 0
3 years ago
Other questions:
  • What is the gcf of 60 and 20
    5·2 answers
  • What is the quotient of 4 and 2/3 divided by 2/3???
    14·2 answers
  • the initial price of a home entertainment system is $34,000. It is discounted by 10% and then by a further 26%. Find the price o
    15·1 answer
  • the sum of three consecutive natural numbers is 156 find the number which is the multiple of 13 out of these numbers​
    6·2 answers
  • Twothirds times fifteen
    11·2 answers
  • Identify the vertex of the graph tell weather it is minimum or maximum
    12·1 answer
  • HELP! The Last option is none of the above
    9·1 answer
  • Solve the equation 12+x=-3(4-x)
    11·1 answer
  • 3. (04.01 MC)
    14·1 answer
  • Sofia can text 40 words in one minute
    13·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!