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3 years ago

Kathy has a coin, what is the probability of flipping three heads in a row? Please choose one answer.

Mathematics

2 answers:

julsineya [31]3 years ago

8 0

Answer:

About 1/8 of a chance or a 12.5% chance

Step-by-step explanation:

See, we already know that the chance of one head is going to be one-half, right?

So we can multiply that fraction by itself by as many times as the coin is flipped in order to get our probability.

So 1/2*1/2*1/2 = 1/8

1/8 is also equal to the decimal .125, and percentages are based of the number one, so as a percentage it should look like this: 12.5%

To be more in depth with it, let's look at this:

Based off this chart, we can also identify that there is a total of 8 possible outcomes, so it is not a coincidence that there is a 1/8 chance of getting heads 3 times.

Katyanochek1 [597]3 years ago

4 0

Answer and Explanation:

The probability of flipping a coin to get heads is 1/2 or 50%. If you multiply this, by how many times you flip it, or 1/2 * 1/2 * 1/2, then you get 1/8 or 12.5% chance that she will get heads 3 times in a row

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Natasha2012 [34]

Answer:

$(1\frac{1}{2},1)$ - False

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$(0,0)$ -- True

$(2\frac{1}{2},3\frac{3}{4})$ -- True

Step-by-step explanation:

The points are

$(1\frac{1}{2},1)$ , $(4,6)$ , $(18,12)$ , $(0,0)$ and $(2\frac{1}{2},3\frac{3}{4})$ ---- missing from the question

Given

$y = 1\frac{1}{2}x$

Required

Determine if each of the points would be on $y = 1\frac{1}{2}x$

To do this, we simply substitute the value of x and of each point in $y = 1\frac{1}{2}x$ .

(a) $(1\frac{1}{2},1)$

In this case;

$x = 1\frac{1}{2}$ and $y = 1$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 1\frac{1}{2}$

$y = \frac{3}{2} * \frac{3}{2}$

$y = \frac{9}{4}$

$y = 2\frac{1}{4}$

The point $(1\frac{1}{2},1)$ won't be on the graph because the corresponding value of y for $x = 1\frac{1}{2}$ is $y = 2\frac{1}{4}$

(b) $(4,6)$

In this case;

$x = 4$

$y = 6$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 4$

$y = \frac{3}{2} * 4$

$y = \frac{3* 4}{2}$

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The point $(4,6)$ would be on the graph because the corresponding value of y for $x = 4$ is $y = 6$

(c) $(18,12)$

In this case:

$x = 18;y = 12$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 18$

$y = \frac{3}{2} * 18$

$y = \frac{3* 18}{2}$

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The point $(18,12)$ wouldn't be on the graph because the corresponding value of y for $x = 18$ is $y = 12$

(d) $(0,0)$

In this case;

$x =0; y = 0$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 0$

$y = 0$

The point $(0,0)$ would be on the graph because the corresponding value of y for $x = 0$ is $y = 0$

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$x = 2\frac{1}{2}$ ; $y = 3\frac{3}{4}$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 2\frac{1}{2}$

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