All categories

1 year ago

What is the probability of tossing two coins in Irvington bus land on heads?

Mathematics

1 answer:

d1i1m1o1n [39]1 year ago

6 0

Sample space = {TT, HH, TH, HT} where T is tail and H is head.
Number of outcomes = 4
The probability of getting two heads on tossing two coins
$= (\frac{1}{4} \times 100)\% \\ = 25\%$

Answer:

D) 25%

Hope you could understand.

If you have any query, feel free to ask.

You might be interested in

one interior angle of a polygon is a right angle and each of the other interior angles is 126°. Calculate the number of sides of

Vikentia [17]

Answer:

There are 7 sides.

Step-by-step explanation:

Let's Use the formula that states that the sum of the angles of an n-sided polygon is given by.

S=(n−2)180

Since we are given that the two angles are right, the angles and each of the remaining angles is 144.

And Therefore, the sum is:

S=90

(n−2)144

(n−2)180

=90

+90

+(n−2)144

⇒(n−2)180

−(n−2)144

=180

⇒(n−2)(180

−144

)=180

⇒(n−2)(36

=180

⇒n−2=

180

⇒n−2=5

⇒n=5+2=7

Hence, the polygon has 7 sides.

6 0

1 year ago

Read 2 more answers

A boat is heading towards a lighthouse, whose beacon-light is 139 feet above the

Novay_Z [31]

The horizontal distance between light house and boat is 1588.78 feet approximately.

The figure is given by,

Here, AB = height of the lighthouse bacon light above the water = 139 feet

Now angle ACB = 5 degree

Let the horizontal distance of light house from the boat = BC = x feet

So by trigonometric function we get,

tan 5 = AB/BC

tan 5 = 139/x

x = 139/tan 5 = 1588.78 (approximately)

Hence the horizontal distance between light house and boat is 1588.78 feet approximately.

To know more about Trigonometric Function refer to:

brainly.com/question/1143565

#SPJ9

5 0

8 months ago

Compute I need help

nataly862011 [7]

$(\sqrt{3}-\sqrt{6}+\sqrt{12}-\sqrt{24})\times \frac{\sqrt{6}}{2} \\ (\sqrt{3}-\sqrt{6}+2\sqrt{3}-\sqrt{24})\times \frac{\sqrt{6}}{2} \\ (\sqrt{3}-\sqrt{6}+2\sqrt{3}-2\sqrt{6})\times \frac{\sqrt{6}}{2} \\ ((\sqrt{3}+2\sqrt{3})+(-\sqrt{6}-2\sqrt{6}))\times \frac{\sqrt{6}}{2} \\ (3\sqrt{3}-3\sqrt{6})\times \frac{\sqrt{6}}{2} \\ \frac{(3\sqrt{3}-3\sqrt{6})\sqrt{6}}{2} \\ \frac{3(\sqrt{3}-\sqrt{6})\sqrt{6}}{2}$