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3 years ago

If you flip a coin 150 times about how many times would you expect to get heads?

2 answers:

8 0

It could go either way... There is a chance of getting more heads than tails or more tails than heads. But if you flip it an even amount of times the best estimation would be that about 50% of the time it will be heads and 50% of the time it will be tails... So you could estimate ABOUT 75 times!

Anit [1.1K]3 years ago

4 0

You cannot have any expectations, statistics cannot apply here, you are as likely, in theory to get 150 heads as none at all as each coin toss is totally independent of the one before and after.
Chaos theory would probably show that you are unlikely to get more or less than a certain number of heads or tails but mathematically there is absolutely no way of predicting a percentage out of any number of tosses.

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What is 10 and 2/3 divided by 8?

postnew [5]

10 and 2/3 = 32/3
32/3 x 8 = 256/3
256/3 = 85.33333333333 repeating

4 0

3 years ago

One of the vertices of an equilateral triangle is on the vertex of a square and two other vertices are on the not adjacent sides

Elina [12.6K]

<h2>Answer:</h2>

<em> The side of the triangle is either 38.63ft or 10.35ft</em>

<h2>Step-by-step explanation:</h2>

This problem can be translated as an image as shown in the Figure below. We know that:

The side of the square is 10 ft.
One of the vertices of an equilateral triangle is on the vertex of a square.
Two other vertices are on the not adjacent sides of the same square.

Let's call:

Since the given triangle is equilateral, each side measures the same length. So:

x: The side of the equilateral triangle (Triangle 1)

y: A side of another triangle called Triangle 2.

That length is the hypotenuse of other triangle called Triangle 2. Therefore, by Pythagorean theorem:

$\mathbf{(1)} \ x^2=100+y^2$

We have another triangle, called Triangle 3, and given that the side of the square is 10ft, then it is true that:

$y+(10-y)=10$

Therefore, for Triangle 3, we have that by Pythagorean theorem:

$(10-y)^2+(10-y)^2=x^2 \\ \\ 2(10-y)^2=x^2 \\ \\ \\ \mathbf{(2)} \ x^2=2(10-y)^2$

Matching equations (1) and (2):

$2(10-y)^2=100+y^2 \\ \\ 2(100-20y+y^2)=100+y^2 \\ \\ 200-40y+2y^2=100+y^2 \\ \\ (2y^2-y^2)-40y+(200-100)=0 \\ \\ y^2-40y+100=0$

Using quadratic formula:

$y_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ \\ y_{1,2}=\frac{-(-40) \pm \sqrt{(-40)^2-4(1)(100)}}{2(1)} \\ \\ \\ y_{1}=37.32 \\ \\ y_{2}=2.68$

Finding x from (1):

$x^2=100+y^2 \\ \\ x_{1}=\sqrt{100+37.32^2} \\ \\ x_{1}=38.63ft \\ \\ \\ x_{2}=\sqrt{100+2.68^2} \\ \\ x_{2}=10.35ft$

<em>Finally, the side of the triangle is either 38.63ft or 10.35ft</em>

5 0

3 years ago

Read 2 more answers

PLEASE ITS MY LAST QUESTION

Serjik [45]

Answer:

$y = - \frac{1}{4} x - 10$

Step-by-step explanation:

The negative reciprocal of 4 is -1/4.

Let's substitute in our values, to find the y-intercept:

$y = - \frac{1}{4} x + c$

$- 11 = - 1 + c$

$c = - 10$

Finally, our full equation is:

$y = - \frac{1}{4} x - 10$

4 0

1 year ago

How do you write 5 39/40 as a decimal?

valkas [14]

First you divide 39/40
you would get 0.975

7 0

3 years ago

Mike has a storage box that is 3 feet long, 2 feet wide, and 1 foot deep. How many 1-foot cubes can the box hold?

alekssr [168]

Answer:

I think the answer is 6.

Step-by-step explanation:

Since the box is 1 foot deep and the cubes are 1 foot, they must sit side by side instead of being stacked up. Therefore you multiply length by width 2x3 and you get 6.

5 0

2 years ago