Can someone help me with numbers 18 and 20? thank you!

A company is trying out a new marketing technique that begin by asking One customer to tell three friends about a product in the

Nostrana [21]

Week 1 : 1 customer tells 3 friends

Week 2: 3 new customers tell 9 friends

Week 3: 9 new customers tell 27 friends

If w represents number of weeks, then (C) represents the model appropriately. Let's see how:

Putting w = 1 , 2 and 3 in each of the options

(A) 3w: 3 for week 1, 6 for week 2, 9 for week 3 ⇒ INCORRECT

(B) w^3: 1 for week 1, 8 for week 2, 27 for week 3 ⇒ INCORRECT

(C) 3^w: 3 for week 1, 9 for week 2, 27 for week 3 ⇒ CORRECT

(D) 3w^3: 3 for week 1, 24 for week 2, 81 for week 3 ⇒ INCORRECT

Hence, the answer is (C)

5 0

3 years ago

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Two metal disks, one with radius r = 2. 50 cm and the other with radius r = 5. 00 cm are welded together and

vova2212 [387]

The total moment of inertia of the two disks will be I = 2.375 × 10-³ Kgm² welded together to form one unit.

<h3>What is moment of inertia?</h3>

Moment of inertia is the quantity expressing a body's tendency to resist angular acceleration, which is the sum of the products of the mass of each particle in the body with the square of its distance from the axis of rotation.

Using the formulas to calculate the moment of inertia of a solid cylinder:

I = ½MR²

Where;

I = moment (kgm²)

M = mass of object (Kg)

R = radius of object (m)

Total moment of inertia of the two disks is expressed as: I = I(1) + I(2)

That is;

I = ½M1R1 + ½M2R2

According to the provided information;

R1 = 2.50cm = 0.025m

M1 = 0.800kg

R2 = 5.00cm = 0.05m

M2 = 1.70kg

I = (½ × 0.800 × 0.025²) + (½ × 0.05² × 1.70)

I = (½ × 0.0005) + (½ × 0.00425)

I = (0.00025) + (0.002125)

I = 0.002375

I = 2.375 × 10-³ Kgm²

Hence The total moment of inertia of the two disks will be I = 2.375 × 10-³ Kgm² welded together to form one unit.

To know more about moment of inertia follow

brainly.com/question/14460640

#SPJ4

4 0

2 years ago

The circumference of the hub cap of a tire is 83.21 centimeters. Find the area of this hub cap. Use

Elena L [17]

Answer:

551 square centimeters

Step-by-step explanation:

Finding the radius (r) :

C = 2πr
83.21 = 2 x 3.14 x r
83.21 = 6.28r
r = 83.21/6.28
r = 13.25 cm

Finding the area :

A = πr²
A = 3.14 x (13.25)²
A = 3.14 x 175.5625
A = 551 square centimeters (nearest whole number)

The area of this hub cap is about 551 square centimeters.

7 0

1 year ago

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A Ferris Wheel 22.0m in diameter rotates once every 12.5s. What is the ratio of a persons apperenet weight to her real weight (a

AnnZ [28]

Acceleration of a passenger is centripetal acceleration, since the Ferris wheel is assumed at uniform speed:
a = omega^2*r

omega and r in terms of given data:
omega = 2*Pi/T
r = d/2

Thus:
a = 2*Pi^2*d/T^2

What forces cause this acceleration for the passenger, at either top or bottom?

At top (acceleration is downward):
Weight (m*g): downward
Normal force (Ntop): upward

Thus Newton's 2nd law reads:
m*g - Ntop = m*a

At top (acceleration is upward):
Weight (m*g): downward
Normal force (Nbottom): upward

Thus Newton's 2nd law reads:
Nbottom - m*g = m*a

Solve for normal forces in both cases. Normal force is apparent weight, the weight that the passenger thinks is her weight when measuring by any method in the gondola reference frame:
Ntop = m*(g - a)
Nbottom = m*(g + a)

Substitute a:
Ntop = m*(g - 2*Pi^2*d/T^2)
Nbottom = m*(g + 2*Pi^2*d/T^2)

We are interested in the ratio of weight (gondola reference frame weight to weight when on the ground):
Ntop/(m*g) = m*(g - 2*Pi^2*d/T^2)/(m*g)
Nbottom/(m*g) = m*(g + 2*Pi^2*d/T^2)/(m*g)

Simplify:
Ntop/(m*g) = 1 - 2*Pi^2*d/(g*T^2)
Nbottom/(m*g) = 1 + 2*Pi^2*d/(g*T^2)

Data:
d:=22 m; T:=12.5 sec; g:=9.8 N/kg;

Results:
Ntop/(m*g) = 71.64%...she feels "light"
Nbottom/(m*g) = 128.4%...she feels "heavy"

7 0

3 years ago

Read 2 more answers

Which of the following are true statements.

Mariulka [41]

Answer:

Second statement is true.

The lengths 7, 40 and 41 can not be sides of a right triangle. The lengths 12, 16, and 20 can be sides of a right triangle.

Step-by-step explanation:

for first part of statement

The lengths 7, 40 and 41 can not be sides of a right triangle.

If the square of long side is equal to the sum of square of other two sides

then the given length can be sides of a right triangle.

Check the given length by Pythagoras Theorem.

$c^{2} =a^{2} +b^{2}$ ----------(1)

Let $c=41$ and $a = 7$ and $b=40$

Put all the value in equation 1.

$41^{2} =7^{2} +40^{2}$

$1681=49+1600$

$1681=1649$

Therefore, the square of long side is not equal to the sum of square of other two sides, So given lengths 7, 40 and 41 can not be sides of a right triangle.

for second part of statement.

The lengths 12, 16, and 20 can be sides of a right triangle.

Check the given length by Pythagoras Theorem.

Let $c=20$ and $a = 12$ and $b=16$

$20^{2} =12^{2} +16^{2}$

$400=144+256$

$400=400$

Therefore, the square of long side is equal to the sum of square of other two sides, So given the lengths 12, 16, and 20 can be sides of a right triangle.

Therefore, The lengths 7, 40 and 41 can not be sides of a right triangle. The lengths 12, 16, and 20 can be sides of a right triangle.

8 0

2 years ago