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3 years ago

7

A customer owns 100 shares of ABC stock in a margin account, valued at $40 per share. The customer sells 3 ABC Jul 40 Calls @ $4

. The maximum potential loss is:

1 answer:

den301095 [7]3 years ago

8 0

Answer:

The maximum potential loss is unlimited

Explanation:

<u>The main reason behind this answer is:</u>

The short calls are covered by the long stock position, however the remaining two short calls are naked. so the maximum potential on short naked calls is unlimited.

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A wave can be best defined as <br><br> P.S pls help

Sidana [21]

Answer:

It should be A. Disturbance that travels through a medium or space, transmitting energy from one point to another.

I hope this helped you :)

4 0

3 years ago

Machmer Hall is 400 m North and 180 m West of Witless.

yan [13]

Answer:

The distance from Witless to Machmer is 438.63 m.

Explanation:

Given that,

Machmer Hall is 400 m North and 180 m West of Witless.

We need to calculate the distance

Using Pythagorean theorem

$D = \sqrt{(d_{m})^2+(d_{w})^2}$

Where, $d_{m}$ =distance of Machmer Hall

$d_{w}$ =distance of Witless

Put the value into the formula

$D = \sqrt{(400)^2+(180)^2}$

$D=438.63\ m$

Hence, The distance from Witless to Machmer is 438.63 m.

5 0

3 years ago

19) A child on a sled starts from rest at the top of a 15.0° slope. If the trip to the bottom takes 15.2 s,

sveticcg [70]

Answer: 288.8 m

Explanation:

We have the following data:

$t=15.2 s$ is the time it takes to the child to reach the bottom of the slope

$V_{o}=0$ is the initial velocity (the child started from rest)

$\theta=15\°$ is the angle of the slope

$d$ is the length of the slope

Now, the Force exerted on the sled along the ramp is:

$F=ma$ (1)

Where $m$ is the mass of the sled and $a$ its acceleration

In addition, if we draw a free body diagram of this sled, the force along the ramp will be:

$F=mg sin \theta$ (2)

Where $g=9.8 m/s^{2}$ is the acceleration due gravity

Then:

$ma=mg sin \theta$ (3)

Finding $a$ :

$a=g sin \theta$ (4)

$a=9.8 m/s^{2} sin(15\°)$ (5)

$a=2.5 m/s^{2}$ (6)

Now, we will use the following kinematic equations to find $d$ :

$V=V_{o}+at$ (7)

$V^{2}=V_{o}^{2}+2ad$ (8)

Where $V$ is the final velocity

Finding $V$ from (7):

$V=at=(2.5 m/s^{2})(15.2 s)$ (9)

$V=38 m/s$ (10)

Substituting (10) in (8):

$(38 m/s)^{2}=2(2.5 m/s^{2})d$ (11)

Finding $d$ :

$d=288.8 m$

6 0

3 years ago

Which force is there between an ice skate and the ice

andrew-mc [135]

The answer is friction

7 0

3 years ago

Read 2 more answers

The submarine emits a pulse of sound to detect other objects in the sea. The sped of sound in sea water is 1500m/s. An echo is r

r-ruslan [8.4K]

Answer:

d = 375 m

Explanation:

The speed of sound is constant in any medium, therefore we can use the uniform motion relationships

v = x / t

x = v t

In this case it indicates that the time since the sound is emitted and received is t = 0.50 s, in this time the sound traveled a round trip distance

x = 2d

2d = v t

d = v t/2

let's calculate

d = 1500 0.5 / 2

d = 375 m

3 0

3 years ago