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Serhud [2]
3 years ago
9

When you're using accounting software why would you use hot keys and shortcuts?

Mathematics
1 answer:
9966 [12]3 years ago
4 0
Hi!

You would use hot keys and shortcuts in an <span>accounting software to quickly do a task such as open a folder or create a new file.

Hot keys and shortcuts are keys or a combination of keys on your keyboard that allow you to do certain things on your computer without using your mouse. Some examples are:
ctrl+c = copy
ctrl+v = paste
ctrl+alt+delete = windows menu
etc...</span>
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The heights of two elevators can be modeled by linear functions. At time t = 0. Elevator A is 16 feet above ground floor.
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Option B: Elevator A descends at a faster rate than elevator B

Option C: Elevator A will reach the ground first

Step-by-step explanation:

Elevator A: -4x+16

Elevator B: -3x +16

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6 in<br> 4 in<br> 4 in.<br> What is the surface area of the right square pyramid?
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The two-way table shows the number of sport utility vehicles with certain features for sale at the car lot. a 4-column table has
Mekhanik [1.2K]

The probability that a randomly selected car with no 4-wheel drive has third-row seats is given by: Option B: 0.4

<h3>How to calculate the probability of an event?</h3>

Suppose that there are finite elementary events in the sample space of the considered experiment, and all are equally likely.

Then, suppose we want to find the probability of an event E.

Then, its probability is given as

P(E) = \dfrac{\text{Number of favorable cases}}{\text{Number of total cases}} = \dfrac{n(E)}{n(S)}

where favorable cases are those elementary events who belong to E, and total cases are the size of the sample space.

<h3>What is chain rule in probability?</h3>

For two events A and B, by chain rule, we have:

P(A \cap B) = P(B)P(A|B) = P(A)P(B|A)

where P(A|B) is probability of occurrence of A given that B already occurred.

For this case, the table given is:

Entries                          4-wheel drive        No 4-wheel drive        Total

Third Row Seats                  18                                12                        30

No Third Row Seats             7                                28                        35

Total                                     25                               40                        65

Let we take

A = event that a randomly selected car has no-4 wheel drive

B = event that a randomly selected car has third row seats

The total ways a car can be selected = 65 = n(S)

Total ways A can happen = n(A) = 40 (from the table).

Similarly, n(B) = 30

P(A) = n(A)/n(S) = 40/65

P(B) = n(B)/n(S) = 30/65

P(A∩B) = n(A∩B)/n(S) = 12/65

As by chain rule, we have:

P(A∩B) = P(A)P(B|A) = P(B)P(A|B)

We need P( A randomly selected car has three seats given that the selected car is with no 4-wheel drive)

which is symbolically P( B | A)

Thus, we use: P(A∩B) = P(A)P(B|A)

or

12/65 = (30/65)(P(B|A))\\\dfrac{12/65}{30/65} = P(B|A)\\\\P(B|A) = \dfrac{12}{30} = \dfrac{2}{5} = 0.4

Thus, the probability that a randomly selected car with no 4-wheel drive has third-row seats is given by: Option B: 0.4

Learn more about probability here:

brainly.com/question/1210781

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2 years ago
Here is my question. will give brainliest
gulaghasi [49]

Answer:

last one

Step-by-step explanation:

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