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3 years ago

The probability of a biased coin landing on heads is 2/7 complete the tree diagram

Mathematics

2 answers:

mamaluj [8]3 years ago

8 0

Answer:

The numbers needed to fill each box are in the image attached

Step-by-step explanation:

The probability of the coin landing on heads is 2/7, so the probability of it landing on tails is 1 - 2/7 = 5/7

The probability of landing heads 2 times is:

P = (2/7) * (2/7) = 4/49

The probability of landing heads and then tails is:

P = (2/7) * (5/7) = 10/49

The probability of landing tails and then heads is:

P = (5/7) * (2/7) = 10/49

The probability of landing tails 2 times is:

P = (5/7) * (5/7) = 25/49

The numbers needed to fill each box are in the image attached.

Mice21 [21]3 years ago

8 0

The first flip numbers will always add up to 1 so you do

1 - 2/7= 5/7

now it gets easy...

The second flip head will be 2/7 because it is a biased dice and the second flip tail will be 5/7

This also applies for the bottom 2: 2/7 head, 5/7 tails

hope this helped :)

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2 years ago

Area of a triangle with points at (-9,5), (6,10), and (2,-10)

Ann [662]

First we are going to draw the triangle using the given coordinates.
Next, we are going to use the distance formula to find the sides of our triangle.
Distance formula: $d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Distance from point A to point B:
$d_{AB}= \sqrt{[6-(-9)]^2+(10-5)^2}$
$d_{AB}= \sqrt{(6+9)^2+(10-5)^2}$
$d_{AB}= \sqrt{(15)^2+(5)^2}$
$d_{AB}= \sqrt{225+25}$
$d_{AB}= \sqrt{250}$
$d_{AB}=15.81$

Distance from point A to point C:
$d_{AC}= \sqrt{[2-(-9)]^2+(-10-5)^2}$
$d_{AC}= \sqrt{(2+9)^2+(-10-5)^2}$
$d_{AC}= \sqrt{11^2+(-15)^2}$
$d_{AC}= \sqrt{121+225}$
$d_{AC}= \sqrt{346}$
$d_{AC}= 18.60$

Distance from point B from point C
$d_{BC}= \sqrt{(2-6)^2+(-10-10)^2}$
$d_{BC}= \sqrt{(-4)^2+(-20)^2}$
$d_{BC}= \sqrt{16+400}$
$d_{BC}= \sqrt{416}$
$d_{BC}=20.40$

Now, we are going to find the semi-perimeter of our triangle using the semi-perimeter formula:
$s= \frac{AB+AC+BC}{2}$
$s= \frac{15.81+18.60+20.40}{2}$
$s= \frac{54.81}{2}$
$s=27.41$

Finally, to find the area of our triangle, we are going to use Heron's formula:
$A= \sqrt{s(s-AB)(s-AC)(s-BC)}$
$A=\sqrt{27.41(27.41-15.81)(27.41-18.60)(27.41-20.40)}$
$A= \sqrt{27.41(11.6)(8.81)(7.01)}$
$A=140.13$

We can conclude that the perimeter of our triangle is 140.13 square units.