How do i solve 2(x+4)=5x-2

A fair coin is flipped seven times. what is the probability of the coin landing tails up at least two times?

GaryK [48]

The required probability of the coin landing tails up at least two times is 15/16.

Given that,
A fair coin is flipped seven times. what is the probability of the coin landing tails up at least two times is to be determined.

<h3>What is probability?</h3>

Probability can be defined as the ratio of favorable outcomes to the total number of events.

Here,
In the given question,
let's approach inverse operation,
The probability of all tails = 1 / 2^7 because there is only one way to flip these coins and get no heads.
The probability of getting 1 head = 7 /2^7
Adding both the probability = 8 / 2^7
Probability of the coin landing tails up at least two times = 1 - 8/2^7
= 1 - 8 / 128
= 120 / 128
= 15 / 16

Thus, the required probability of the coin landing tails up at least two times is 15/16.

Learn more about probability here:

brainly.com/question/14290572

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7 0

1 year ago

Joe flipped a coin 100 times. It landed heads

gayaneshka [121]

Answer:

48%

Step-by-step explanation:

48 out of 100 = 48%

:D

6 0

3 years ago

Read 2 more answers

Which element of variation would not be affected by adding the data value 15 to the data set {3, 5, 6, 8, 9, 10, 13, 14}

Andrei [34K]

It is 13,14 because it goes up two then one one then two again which will then mess up the rotation

8 0

3 years ago

Find the horizontal and vertical asymptotes of f(x). f(x) equals = StartFraction 6 x Over x plus 2 EndFraction 6x x+2 Find the

miss Akunina [59]

Answer:

The horizontal asymptote can be described by the line y = 6

The vertical asymptote can be described by the line x = -2

Step-by-step explanation:

* Lets the meaning of vertical and horizontal asymptotes

- Vertical asymptotes are vertical lines which correspond to the zeroes

of the denominator of a rational function

- A horizontal asymptote is a y-value on a graph which a function

approaches but does not actually reach

- If the degree of the numerator is less than the degree of the

denominator, then there is a horizontal asymptote at y = 0

- If the degree of the numerator is greater than the degree of the

denominator, then there is no horizontal asymptote

- If the degree of the numerator is equal the degree of the denominator,

then there is a horizontal asymptote at y = leading coefficient of the

numerator ÷ leading coefficient of the denominator

* Lets solve the problem

∵ $f(x)=\frac{6x}{x+2}$

∵ The numerator is 6x

∵ The denominator is x + 2

∴ The numerator and the denominator have same degree

∵ The leading coefficient of the numerator is 6

∵ The leading coefficient of the denominator is 1

∴ There is a horizontal asymptote at y = 6/1

∴ The horizontal asymptote can be described by the line y = 6

- Put the denominator equal zero to find its zeroes

∵ The denominator is x + 2

∴ x + 2 = 0

- Subtract 2 from both sides

∴ x = -2

∴ The vertical asymptote can be described by the line x = -2

6 0

3 years ago

Read 2 more answers

when a cannonball is fired, the equation of its pathway can be modeled by h=-16t squared + 128t. find the time it will take for

Rashid [163]

At the time the cannonball reaches the ground, the vertical height of the cannonball from the ground is 0, i.e. h = 0.

Thus, the time it will take the cannonball to reach the ground is given by

$-16t^2+128t=0 \\ \\ \Rightarrow t(-16t+128)=0 \\ \\ \Rightarrow t=0\ or\ -16t+128=0 \\ \\ \Rightarrow16t=128 \\ \\ \Rightarrow t= \frac{128}{16} =8$

Therefore, it will take the cannonball 8 seconds to reach the ground.

5 0

3 years ago