Please add a number line if you can

1. A weighted coin is biased so that a head is twice as likely to occur as a tail. If the coin is tossed 4 times, what is the pr

ohaa [14]

Answer:

80/81

Step-by-step explanation:

If a head is twice as likely to occur as a tail, then the probability of getting heads is 2/3 and the probability of getting tails is 1/3.

The probability of getting at least 1 head involves 4 scenarios:

1) 1 Head and 3 Tails

2) 2 Heads and 2 Tails

3) 3 Heads and 1 Tail

4) 4 Heads

Instead of calculate all these scenarios, you could calculate the opposite scenario: 4 Tails. The sum of all possible scenarios is 1, so:

P(at least one head) + P(no heads) = 1

Then, P(at least one head) = 1 - P(no heads)

The probability of 4 tails is:

P(no heads) = P(TTTT) = (1/3)(1/3)(1/3)(1/3)=1/81

Then, P(at least one head) = 1 - 1/81=80/81

7 0

3 years ago

circle q is centered at the origin with radius r point p(xy)lies on circle q make conjecture how can you find an equation relati

babunello [35]

Answer:

[tex]x^{2} +y^{2}=r^{2}[/tex]

Step-by-step explanation:

We know that the center-radius form of the circle equation is in the format (x – h)2 + (y – k)2 = r2, with the center being at the point (h, k) and the radius being "r".

As given the center of circle q is origin ie,(0,0);

$h=0 and k=0$

Now substitute the point P in the equation of the circle as it lies on it.

$(x-0)^{2}+(y-0)^{2}=r^{2}$

$x^{2}+y^{2}=r^{2}$

8 0

3 years ago

1) Para que valores reais de x o ponto P (3x -6, 2x +4) pertence ao 4º quadrante?

skelet666 [1.2K]

Answer:

There are no real values of x for point P to belong to the 4 quadrant

Step-by-step explanation:

<u><em>The question in English is</em></u>

To what real values of x does the point P (3x -6, 2x +4) belong to the 4th quadrant?

we know that

A point in the fourth Quadrant has the x-coordinate positive and the y-coordinate negative

we have the point

P (3x-6, 2x+4)

$3x-6 > 0$ ----> inequality A ( x-coordinate must be positive)

$2x+4 < 0$ ---> inequality B ( y-coordinate must be negative)

Solve Inequality A

$3x-6 > 0$

$3x > 6$

$x > 2$ -----> (2,∞)

Solve Inequality B

$2x+4 < 0$

$2x < -4$

$x < -2$ ----> (-∞,-2)

The solution of the system is

(-∞,-2) ∩ (2,∞)

therefore

The system has no solution

There are no real values of x for point P to belong to the 4 quadrant

3 0

4 years ago

How do I simplify 3⁶ x 3⁵

elena55 [62]

Answer:

$3^{11}$

Step-by-step explanation:

Using the rule of exponents

$a^{m}$ × $a^{n}$ ⇔ $a^{(m+n)}$ thus

$3^{6}$ × $3^{5}$ = $3^{(6+5)}$ = $3^{11}$

6 0

4 years ago

Read 2 more answers

Express the quantity of the Cayennepepper im lemon juice in form one

Nesterboy [21]

Answer:

HUH

Step-by-step explanation:

8 0

3 years ago