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3 years ago

What is the experamental probability that the coin lands on heads?

Mathematics

1 answer:

sammy [17]3 years ago

5 0

Answer:

The experamental probability that the coin lands on head is 50 %

Step-by-step explanation:

Given:

Experiment:

A coin is Toss

Let the Sample Space be 'S' that is total number of outcomes for a coin has been tossed = { Head, Tail }

∴ n ( S ) = 2

Let A be the event of getting a Head on tossing a coin i.e { Head }

∴ n( A ) = 1

Now,

$\textrm{Probability} =\frac{\textrm{outcomes for the experiment}}{\textrm{total number of outcomes}}$

Substituting the values we get

$P(A) = \frac{n(A)}{n(S)} \\\\P(A) = \frac{1}{2} \\\\P(A) = 0.5\\\\P(A) = 50\%$

The experamental probability that the coin lands on head is 50 %

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Whats the product of 0.02 x0.3?PLEASE ANSWER WITH STEPS PLEASEEEE,ITS URGENT

motikmotik

Answer:

0.006

Step-by-step explanation:

This is the answer because:

Short Way:

1) We know 3 x 2 is 6

2) When we multiply with zeros, how many ever zeros there are in the factors together, should be in the answer. In this case, in the answer we should have 3 zeros, one before the decimal point and two after the decimal point, because in 0.02 there are two zeros while in 0.3 there is only one zero.

3) Finally, the answer would be 0.006

Long Way:

1) You can do long multiplication to solve for the answer.

2) 0.02

× 0.3

Is 0.006 because 3 x 2 is 6, and then 3 x 0 is 0 and then 3 x 0 is 0.

006

+0000

0.006

Remember to put a decimal point.

Hope this helps!

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4 years ago

Hey answer this those who know?

Makovka662 [10]

Answer:

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3 years ago

The sum of the diagonals of a rhombus is 5√2.

Alex

Greetings!
Let ABCD be a rhombus
AC + BD = 5√2 cm

Area of ABCD = Ar△ABD + Ar △BCD
$= \frac{1}{2} \times BD \times AO + \frac{1}{2} \times BD \times OC$
$= \frac{1}{2} BD (AO + OC)$
$\frac{1}{2} BD \times AC$

$So, \frac{ BD \times AC}{2} = 4 \: cm {}^{2}$
$BD \times AC = 8$
$Now, AC + \: BD = 5 \sqrt{2}$

Squaring both sides, we get
$AC {}^{2} + BD {}^{2} + 2 AC.BD =50$
$AC {}^{2} + BD {}^{2} + 2 \times 8 = 50$
$AC {}^{2} + BD {}^{2} = 50 - 16$
$AC {}^{2} + BD {}^{2} = 34$

In △AOB, we have
$OA {}^{2} + OB {}^{2} =AB {}^{2}$
$( \frac{ AC }{2} ) {}^{2} + ( \frac{ BD}{2} ) {}^{2} = AB {}^{2}$
$\frac{ AC {}^{2} } {4} + \frac{ BD {}^{2} }{4} = AB {}^{2}$
$AC {}^{2} + BD {}^{2} = 4 AB {}^{2}$
$34 = 4 AB {}^{2}$

Square rooting both sides
$\sqrt{34} = 2 AB$
$Perimeter = 4 \: AB \\ = 2 \times 2 \: AB \\ = 2 \: \times \sqrt{34 } \\ = 2 \sqrt{34 \: } units$ .

Hope it helps!

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