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2 years ago

A basketball team plays 3 games in a holiday tournament. According to the tree diagram, how many outcomes are possible?

Mathematics

2 answers:

kati45 [8]2 years ago

8 0

Answer:

Hence, the possible number of outcomes are:

Option: d ( d. 8)

Step-by-step explanation:

It is given that:

A basketball team plays 3 games in a holiday tournament.

( We know that:

Tree diagrams display all the possible outcomes of an event. Each branch in a tree diagram represents a possible outcome )

The possible number of outcomes that are possible are:

Let the team be denoted by 'A'

so, the outcome of the game may be winning(W) or losing(L) the game.

dybincka [34]2 years ago

6 0

Answer:

Option d. 8 is the correct option.

Step-by-step explanation:

A basketball team plays 3 games in a holiday tournament.

Let the chance of winning of the team be represented as W and losing by L

As per tree diagram attached we can see

In the first game

Number possibilities will be 2 either win or lose

After second games

Number of possibilities will be 4 [(L L), (L W), (W L), (W W)]

After third game

Number of possibilities will be 8 [ (L L), (L W), (W L), (W W), (L L), (L W), (W L), (W W)]

So total outcomes which are possible are 8.

Option d. 8 is the correct option.

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2 years ago

Find the real numbers x and y if -3+ix^2y and x^2+y+4i are conjugate of each other. Pls solve with the steps

Firdavs [7]

ANSWER
x = ±1 and y = -4.
Either x = +1 or x = -1 will work

EXPLANATION
If -3 + ix²y and x² + y + 4i are complex conjugates, then one of them can be written in the form a + bi and the other in the form a - bi. In other words, between conjugates, the imaginary parts are same in absolute value but different in sign (b and -b). The real parts are the same

For -3 + ix²y
⇒ real part: -3
⇒ imaginary part: x²y

For x² + y + 4i
⇒ real part: x² + y (since x, y are real numbers)
⇒ imaginary part: 4

Therefore, for the two expressions to be conjugates, we must satisfy the two conditions.

Condition 1: Imaginary parts are same in absolute value but different in sign. We can set the imaginary part of -3 + ix²y to be the negative imaginary part of x² + y + 4i so that the

x²y = -4 ... (I)

Condition 2: Real parts are the same

x² + y = -3 ... (II)

We have a system of equations since both conditions must be satisfied

x²y = -4 ... (I)
x² + y = -3 ... (II)

We can rearrange equation (II) so that we have

y = -3 - x² ... (II)

Substituting into equation (I)

x²y = -4 ... (I)
x²(-3 - x²) = -4
-3x² - x⁴ = -4
x⁴ + 3x² - 4 = 0
(x² + 4)(x² - 1) = 0
(x² + 4)(x-1)(x+1) = 0

Therefore, x = ±1.
Leave alone (x² + 4) as it gives no real solutions.

Solve for y:

y = -3 - x² ... (II)
y = -3 - (±1)²
y = -3 - 1
y = -4

So x = ±1 and y = -4. We can confirm this results in conjugates by substituting into the expressions:

-3 + ix²y
= -3 + i(±1)²(-4)
= -3 - 4i

x² + y + 4i
= (±1)² - 4 + 4i
= 1 - 4 + 4i
= -3 + 4i

They result in conjugates

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Answer:

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