A law firm is going to designate associates and partners to a big new case. The daily rate charged to the client for each associ

Gerry has 25 more basketball cards than he does baseball cards. Step-by-step explanation: Gerry owns a total of 50 sports trading cards. Among these, one-fifth are baseball cards, meaning he has 10 baseball cards. One-tenth of his collection consists of football cards, so he has 5 football cards. The remainder are basketball cards, leading to the conclusion that he has 35 basketball cards. The difference in quantity between his basketball and baseball cards is 25.

Answer:

Step-by-step explanation:

There are 15 antennas in total.

Out of these, 3 are defective.

This means that 12 antennas are functioning: 15-3=12.

To ensure that no two defective antennas are adjacent, we need to have only one defective at a time placed between the functional ones.

So,

We align the 13 functional antennas, then look for the spaces where the defective antennas can fit

__G __ G __ G __ G __ G __G __ G __ G __ G __ G __ G __ G __G __

Each gap represented by an underscore indicates a possible location for a defective antenna, allowing for just one per space.

Consequently, there are 14 potential spots for the defective antennas. With 3 defectives, we are dealing with a combinatorial arrangement.

ⁿCr= n!/(n-r)!r!

The total number of arrangements possible is

14C3=14!/(14-3)!3!

14C3=14×13×12×11!/11!×3×2

14C3=14×13×12/6

That gives us 364 distinct ways to arrange them.

Answer:

Step-by-step explanation:

The probability that a mosquito lands on your neck per second is 0.5

The chance of it biting once it lands is 0.2

The likelihood that it does not bite after landing is 0.8

Therefore, the probability of a bite in one second is

This translates to a 1/100 likelihood of being bitten in just one second

Over the course of 100 seconds, the expected number of bites is

The anticipated interval between bites is 100

The corresponding Venn Diagram for this issue is depicted below

P(E|F) and P(F|E) denote the conditional probabilities.

P(E|F) is calculated as P(E∩F) ÷ P(F) = ¹/₂ ÷ ¹/₂ = 1
P(F|E) is determined by P(F∩E) ÷ P(E) = ¹/₂ ÷ ¹/₂ = 1