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3 years ago

What is the experimental probability of a coin toss results in two heads showing?

Mathematics

2 answers:

lorasvet [3.4K]3 years ago

8 0

Answer:

1 in 4 chance

25% chance

Liono4ka [1.6K]3 years ago

6 0

Answer:

25%

(1/4.)

Step-by-step explanation:

So, to get heads twice in a row in two coin tosses it would be 25%. The odds of getting a heads with one coin toss is 50% (1/2.) Every time you do another coin toss the odds are halfed again. So half of 1/2 is 1/4 (think about it as like 50 divided by 2, so 25%.)

Hope this helps and have a nice day! Sorry I'm not great at explaining!

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Solve and reduce to lowest terms: x = Please don't make it to long

dusya [7]

Answer:

2/11

Step-by-step explanation:

6 = 2 x 3

6/11 x 1/3

= ( 6 x 1 )/( 11 x 3 )

= ( 2 x 3 x 1 )/( 11 x 3 )

Cancel 3 in both numerator and denominator.

= ( 2 x 1 )/( 11 )

= 2/11

4 0

2 years ago

25^x = 1 find the value of x

andrew11 [14]

Answer:

x=0

Step-by-step explanation:

Anything to the power of 0 = 1

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3 years ago

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Viktor [21]

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3 years ago

QS¯¯¯¯¯¯¯ bisects ∠PQR . If m∠PQS = 5x and m∠RQS = 2x + 6, then what is m∠PQR ? m∠PQR is

soldi70 [24.7K]

Answer:

$PQR = 20$

Step-by-step explanation:

Given

Bisector: QS

$PQS = 5x$

$RQS = 2x + 6$

Required

Determine PQR

Since PQR is bisected to PQS and RQS, we have that

$PQS = RQS$

Substitute expressions for PQS and RQS

$5x = 2x + 6$

Collect Like Terms

$5x - 2x = 6$

$3x = 6$

Divide both sides by 3

$x = 2$

Solving for PQR;

$PQR = PQS + RQS$

$PQR = 5x + 2x + 6$

$PQR = 7x + 6$

Substitute 2 for x

$PQR = 7 * 2 + 6$

$PQR = 14 + 6$

$PQR = 20$

4 0

3 years ago

Find side lengths C and H

Nuetrik [128]

Answer:

$\displaystyle c=\frac{2\sqrt{3}}{3}\\\\\displaystyle h=7\sqrt{2}\\$

Step-by-step explanation:

Right Triangles

A right triangle can be identified by the fact it has an internal angle of 90°. In a right triangle, the trigonometric ratios stand.

Let's consider the triangle to the left. We need to calculate side c, one of the legs of the triangle. We can use the angle adjacent to it (60°) or the angle opposite to it (30°) with the appropriate trigonometric ratio.

We'll use the adjacent angle, and

$\displaystyle tan60^o=\frac{2}{c}$

Solving for c

$\displaystyle c=\frac{2}{tan60^o}=\frac{2}{\sqrt{3}}$

Rationalizing

$\displaystyle c=\frac{2}{\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}}=\frac{2\sqrt{3}}{3}$

Now for the triangle to the right. The side h is the hypotenuse. Again, any of the two angles can be used (though they are equal, for it's an isosceles triangle). For any of them it is true that

$\displaystyle sin45^o=\frac{7}{h}$

Solving for h

$\displaystyle h=\frac{7}{sin45^o}=\frac{7}{\frac{\sqrt{2}}{2}}=\frac{14}{\sqrt{2}}$

Rationalizing

$\displaystyle h=\frac{14}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{14\sqrt{2}}{2}=7\sqrt{2}$

3 0

3 years ago