Ask question

Ask question

All categories

3 years ago

10

Companies whose stocks are listed in a stock exchange have their company name represented by either three, four, or five letters

(repetition of letters is allowed). What is the maximum number of companies that can be listed on the stock exchange?

1 answer:

nlexa [21]3 years ago

5 0

Answer:

83330 Companies

Step-by-step explanation:

The answer is

26C3 + 26C4 + 26C5

= 2600 + 14950 + 65780

= 83330

You might be interested in

A magician asks two volunteers to each draw a card from a standard deck of cards. What is the probability that the first card is

mars1129 [50]

Answer:

6.37%

Step-by-step explanation:

A deck of cards have 52 cards. There are 4 suits of 13 card each, those suits are Hearts, Clubs, Diamond and Spades.

The probability that the first card is a heart is:

$P(Hearts) = \frac{13}{52}$

Now, the probabily that the second card is a diamond is:

$P(Diamond | First card is Heart) = \frac{13}{51}$

The Probability that the first card is a heart and the second one a diamond is given by:

$P(Hearts)$ × $P(Diamond | First card is Heart)$

$= \frac{13}{52}\frac{13}{51}$

$= 6.37$ %

4 0

3 years ago

Read 2 more answers

While training for a marathon, Jeff wants to increase the number of miles he runs each day. On the first day of training, Jeff r

user100 [1]

Answer:

The series is 5 , 6 , 7 , 8 , 9 , 10 , 11

Step-by-step explanation:

* Lets revise the arithmetic series

- In the arithmetic series there is a constant difference between

each two consecutive numbers

- Ex:

# 2 , 5 , 8 , 11 , ………………………. (constant difference is 3)

# 5 , 10 , 15 , 20 , ………………………… (constant difference is 5)

# 12 , 10 , 8 , 6 , …………………………… (constant difference is -2)

* General term (nth term) of an Arithmetic series:

- If the first term is a and the common diffidence is d, then

U1 = a , U2 = a + d , U3 = a + 2d , U4 = a + 3d , U5 = a + 4d

- So the nth term is Un = a + (n – 1)d, where n is the position of the

number in the series

* Lets solve the problem

- Jeff wants to increase the number of miles he runs each day

∴ He will add the initial value by constant number each day

- He plans on increasing the number of miles he runs a day by 1

∴ The constant value is 1 mile

- On the first day of training, Jeff runs 5 miles

∴ The first value is 5 miles

∴ The series is arithmetic

∵ a = 5 , d = 1

- He do that for the remainder of the week

∵ The week has 7 days

∴ The series has 7 terms

∵ The rule of the series is Un = a + (n - 1)d

∵ a = 5 and d = 1

∴ Un = 5 + (n - 1)(1)

∴ Un = 5 + n - 1

∴ Un = 4 + n ⇒ n is the position of the number

- Substitute n from 1 to 7 to find the series

∴ The series is 5 , 6 , 7 , 8 , 9 , 10 , 11

6 0

3 years ago

Find the standard equation of a sphere that has diameter with the end points given below. (3,-2,4) (7,12,4)

DiKsa [7]

Answer:

The standard equation of the sphere is $(x-5)^{2} + (y-5)^{2} + (z-4)^{2} = 53$

Step-by-step explanation:

From the question, the end point are (3,-2,4) and (7,12,4)

Since we know the end points of the diameter, we can determine the center (midpoint of the two end points) of the sphere.

The midpoint can be calculated thus

Midpoint = $(\frac{x_{1} + x_{2} }{2}, \frac{y_{1} + y_{2} }{2}, \frac{z_{1} + z_{2} }{2})$

Let the first endpoint be represented as $(x_{1}, y_{1}, z_{1})$ and the second endpoint be $(x_{2}, y_{2}, z_{2})$ .

Hence,

Midpoint = $(\frac{x_{1} + x_{2} }{2}, \frac{y_{1} + y_{2} }{2}, \frac{z_{1} + z_{2} }{2})$

Midpoint = $(\frac{3 + 7 }{2}, \frac{-2+12 }{2}, \frac{4 + 4 }{2})$

Midpoint = $(\frac{10 }{2}, \frac{10}{2}, \frac{8 }{2})\\$

Midpoint = $(5, 5, 4)$

This is the center of the sphere.

Now, we will determine the distance (diameter) of the sphere

The distance is given by

$d = \sqrt{(x_{2} - x_{1})^{2} +(y_{2} - y_{1})^{2} + (z_{2}- z_{1})^{2} }$

$d = \sqrt{(7 - 3)^{2} +(12 - -2)^{2} + (4- 4)^{2}$

$d = \sqrt{(4)^{2} +(14)^{2} + (0)^{2}$

$d = \sqrt{16 +196 + 0$

$d =\sqrt{212}$

$d = 2\sqrt{53}$

This is the diameter

To find the radius, r

From $Radius = \frac{Diameter}{2}$

$Radius = \frac{2\sqrt{53} }{2}$

∴ $Radius = \sqrt{53}$

r = $\sqrt{53}$

Now, we can write the standard equation of the sphere since we know the center and the radius

Center of the sphere is $(5, 5, 4)$

Radius of the sphere is $\sqrt{53}$

The equation of a sphere of radius r and center $(h,k,l)$ is given by

$(x-h)^{2} + (y-k)^{2} + (z-l)^{2} = r^{2}$

Hence, the equation of the sphere of radius $\sqrt{53}$ and center $(5, 5, 4)$ is

$(x-5)^{2} + (y-5)^{2} + (z-4)^{2} = \sqrt{(53} )^{2}$

$(x-5)^{2} + (y-5)^{2} + (z-4)^{2} = 53$

This is the standard equation of the sphere

6 0

3 years ago

Is the relationship in this picture a function<br>if so why?

navik [9.2K]

Answer:

PLEASE BE CAREFUL NightmareMech14 IS A PREDATOR HE ASKS FOR VIDEOS OF GIRLS AND TRYS TO GET THEM TO EMAIL HIM DONT GIVE HIM ANY INFORMATION STAY SAFE COPY AND PASTE THIS EVERYWHERE

Step-by-step explanation:

3 0

2 years ago

Find all points having an x coordinate of 5 whose distance from the point (2,2) is 5.

stepan [7]

Answer:

(5,6) , (5,-2)

Step-by-step explanation:

Let the required point be P(5,y).

Its distance from (2,2) is 5 units.

So, $(5-2)^{2}$ + $(y-2)^{2}$ = 25.

So, $3^{2}$ + $(y-2)^{2}$ = 25.

9 + $(y-2)^{2}$ = 25.

$(y-2)^{2}$ = 16

Two cases are possible for now

case 1 : y-2 = 4 .

y = 6

required point will be (5,6)

case 2 : y - 2 = -4.

y = -2.

required point will be (5,-2).

Two points are possible : (5,6) , (5,-2).

5 0

3 years ago