All categories

2 years ago

Select the correct comparison

Mathematics

1 answer:

Anastaziya [24]2 years ago

5 0

Answer:

A. typical value and the spread are both greater in set A.

You might be interested in

What is the value of x in the equation −6 + x = −1? (5 points)

Mice21 [21]

Answer:

Step-by-step explanation:

Add 6 to both sides. The answer is 5.

5 0

2 years ago

Read 2 more answers

A North American river otter wants to cross the North Saskatchewan river. It can swim at a speed of 2.75 m/s. In that location,

insens350 [35]

Answer:

Step-by-step explanation:

time to cross will be the distance divided by the velocity in that direction.

a) t = d/v = 45/2.75 = 16.363636... 16.4 s

In the same time that the otter is swimming cross stream, the stream is also moving perpendicularly at its own pace.

b) d = vt = 2.5(16.36) = 40.9090... 40.9 m

As velocity is a vector, An observer on the bank would see the velocity as the vector addition of the cross stream and down stream velocities

c) v = √(2.75² + 2.5²) = 3.716517... 3.72 m/s magnitude

θ = arctan2.5/2.75 = 42.2736... 42.3° downstream of a line straight across

7 0

2 years ago

Um ok i need help pls give correct answers

vesna_86 [32]

-a) function 2, function 4
-b) function 4
-c) function 2

4 0

2 years ago

Read 2 more answers

A hoop, a uniform solid cylinder, a spherical shell, and a uniform solid sphere are released from rest at the top of an incline.

Serjik [45]

Answer:

Step-by-step explanation:

Given

Hoop, Uniform Solid Cylinder, Spherical shell and a uniform Solid sphere released from Rest from same height

Suppose they have same mass and radius

time Period is given by

$t=\sqrt{\frac{2h}{a}}$ ,where h=height of release

a=acceleration

$a=\frac{g\sin \theta }{1+\frac{I}{mr^2}}$

Where I=moment of inertia

a for hoop

$a=\frac{g\sin \theta }{1+\frac{mr^2}{mr^2}}$

$a=\frac{g\sin \theta }{2}$

a for Uniform solid cylinder

$a=\frac{g\sin \theta }{1+\frac{mr^2}{2mr^2}}$

$a=\frac{2g\sin \theta }{3}$

a for spherical shell

$a=\frac{g\sin \theta }{1+\frac{2mr^2}{3mr^2}}$

$a=\frac{3g\sin \theta }{5}$

a for Uniform Solid

$a=\frac{g\sin \theta }{1+\frac{2mr^2}{5mr^2}}$

$a=\frac{5g\sin \theta }{7}$

time taken will be inversely proportional to the square root of acceleration

$t_1=k\sqrt{2}=1.414k$

$t_2=k\sqrt{\frac{3}{2}}=1.224k$

$t_3=k\sqrt{\frac{5}{3}}=1.2909k$

$t_4=k\sqrt{\frac{7}{5}}=1.183k$

thus first one to reach is Solid Sphere

second is Uniform solid cylinder

third is Spherical Shell

Fourth is hoop

3 0

3 years ago

Fill in the blanks

brilliants [131]

Answer:

1. 100 which Is the easier one

2. 5

5 0

2 years ago