Generate Quiz 69AB10FA6E75E

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1 / 10

If (n + 2)! = 2550 (n!). find n:

n+1 = 50

n = 49

or reverse the solution

n+1 = 50

n = 49

or reverse the solution

2 / 10

The number of significant numbers which can be formed by using any number of the digits 0, 1, 2, 3, 4 but using each not more than once in each number is

260

356

410

THE RIGHT ANSWER (A)

Five times the factor!

= 120

THE RIGHT ANSWER (A)

Five times the factor!

= 120

3 / 10

How many seven digit telephone number can be formed:

10^2

9*10^6

10p7

4 / 10

Shazia was to choose a password for her bank account. Her favorite digits from 0 – 9 are 0, 1, 3, 4, 6, 8. The problem was that the password could only be of 4 digits. How many different passwords could Shazia choose from?

120

240

360

480

“360” is the right response.

It can be resolved as follows:

When n is the total number of options and r is the number of required options, the notation for permutation is nPr. Shazia can choose from a total of six digits, but she only needs four to create the password.

6P4 = 360

“360” is the right response.

It can be resolved as follows:

6P4 = 360

5 / 10

Factorial of -2 =

-362880

362880

40320

None

RIGHT ANSWER (D).For non-negative integers n, the factorial function, represented by n!, is defined. Up to and including n, it is the product of all positive integers.For negative integers, however, the factorial function is not defined. Therefore, the factorial of −2 does not exist.

6 / 10

A bag contains 3 white, 4 black and 2 red balls, if 2 balls are drawn at random, then the probability that both the balls are white is

1/18

1/12

1/36

None of these

THE RIGHT ANSWER (B)

3+4+2=93+4+2=9 is the total number of balls in the bag.

There are three white balls.

There are a total of nine ways to draw two balls: Combinations can be used to calculate this. Since the order in which the balls are drawn is irrelevant, we employ combinations.

The sum of all the ways is (92)=9!2!(9−2)!=9×82×1=36The sum of the ways is (29), which equals 2!(9-2)!9! = 2×19×8 = 3.

Number of favorable outcomes (drawing two white balls): (32)(23) gives the number of ways to choose two white balls out of the three in the bag.

How many ways are there to select two white balls? (32)=3!2!(3−2)!=3×22×1=3. There are 23 ways to choose two white balls, which is equal to 2!(3−2)!3! = 2×13×2 = 3

Finding the probability: The ratio of the number of favorable outcomes to the total number of outcomes is the probability of drawing two white balls.

Total number of outcomes = P(both white)3/36 = 1/12 is the number of favorable outcomes.

Therefore, there is a 1/2 chance that both balls drawn will be white.

THE RIGHT ANSWER (B)

3+4+2=93+4+2=9 is the total number of balls in the bag.

There are three white balls.

There are a total of nine ways to draw two balls: Combinations can be used to calculate this. Since the order in which the balls are drawn is irrelevant, we employ combinations.

The sum of all the ways is (92)=9!2!(9−2)!=9×82×1=36The sum of the ways is (29), which equals 2!(9-2)!9! = 2×19×8 = 3.

Number of favorable outcomes (drawing two white balls): (32)(23) gives the number of ways to choose two white balls out of the three in the bag.

How many ways are there to select two white balls? (32)=3!2!(3−2)!=3×22×1=3. There are 23 ways to choose two white balls, which is equal to 2!(3−2)!3! = 2×13×2 = 3

Finding the probability: The ratio of the number of favorable outcomes to the total number of outcomes is the probability of drawing two white balls.

Total number of outcomes = P(both white)3/36 = 1/12 is the number of favorable outcomes.

Therefore, there is a 1/2 chance that both balls drawn will be white.

7 / 10

The number of diagonals of a 5-sided polygon is:

The following formula provides the number of diagonals in a polygon with n sides:

n * (n – 3) / 2

For a pentagon, or five-sided polygon:

5 * (5 – 3) / 2 = 5.

Thus, (a) 5 is the right response.

The following formula provides the number of diagonals in a polygon with n sides:

n * (n – 3) / 2

For a pentagon, or five-sided polygon:

5 * (5 – 3) / 2 = 5.

Thus, (a) 5 is the right response.

8 / 10

Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls, if each selection consists of 3 balls of each colours:

3000

1000

2000

4000

9 / 10

The sample space for tossing a coin once is

(H)

(T)

(H, H)

(H, T)

Either head or tail will be obtained, resulting in a head-tail sample space.

10 / 10

If a coin is tossed 4 times in a row, what is the probability that at least one tail shows up?

9/16

11/16

13/16

15/16

Answer:

The right response is d) 15/16.

The following outcomes are possible when a coin is tossed four times:

HHHH, HHHH, HHHT, HHTH, HTTH, THTH

The probability of receiving all heads (no tails) is equal to the probability of receiving at least one tail.

According to the calculation in the previous question, the likelihood of receiving no tails is 1/16.

The complement of 1/16, or 15/16, is the likelihood of receiving at least one tail. (The likelihood that an event won’t happen)

Answer:

The right response is d) 15/16.

The following outcomes are possible when a coin is tossed four times:

HHHH, HHHH, HHHT, HHTH, HTTH, THTH

The probability of receiving all heads (no tails) is equal to the probability of receiving at least one tail.

According to the calculation in the previous question, the likelihood of receiving no tails is 1/16.

The complement of 1/16, or 15/16, is the likelihood of receiving at least one tail. (The likelihood that an event won’t happen)

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