<u>Problem</u>
A bag contains 6 blue marbles, 10 red marbles, and 9 green marbles. If two marbles are drawn at random without replacement, what is the probability that two red marbles are drawn?
<u>Work </u>
Probability = no. of favorable outcomes /total no. of outcomes
Probability of getting a blue marble=5/5+6+9=5/20
Probability of getting a red marble=6/20−1=6/19
5/20×6/19
<u>Answer</u>
3/20
So it is C.
Answer:
0
Step-by-step explanation:
Answer:
To develop a molecular clock, you need to find which of the following?
a sequence of molecules
the rate at which changes occur in a type of molecule
how much total change has occurred in a type of molecule from two different species
how many molecules a species has
Step-by-step explanation:
To develop a molecular clock, you need to find which of the following?
a sequence of molecules
the rate at which changes occur in a type of molecule
how much total change has occurred in a type of molecule from two different species
how many molecules a species has
Answer:
can u convert this to miles?
Step-by-step explanation:
Answer:
C
Step-by-step explanation:
To make it easy let's start by organizing our information :
- AC=12 AND BD=8
- ABCD is a rhombus
- K and L are the midpoints of sides AD and CD
- we notice that the rhombus ABCD is divided into four right triangles
What do you think of when you hear a right triangle ?
- The pythagorian theorem !
AC and BD are khown so let's focus on them .
If we concentrated we can notice that AB and BD are cossing each other in the midpoints . why ?
Simply because they are the diagonals of a rhombus .
ow let's apply the pythagorian theorem :
- (AC/2)² + (BD/2)² = BC²
- 6²+4²=52
- BC²= 52⇒
=BC
Now we khow that : AB=BC=CD=AD=
This isn't enough . Let's try to figure out a way to calculate the length of KL wich is the base of the triangle
- KL is parallel to AC
- k is the midpoint of AD and L of DC
I smell something . yes! Thales theorem
- KL/AC=DL/DC=DK/AD WE4LL TAKE OLY ONE
- KL/12=
/2*
- KL/12=1/2⇒ KL=6
Now we have the length of the base kl
Now the big boss the height :
- notice that you khow the length of KL
- BD crosses kl from its midpoint and DL =
/2
What I want to do is to apply the pythgorian thaorem to khow the lenght of that small part that is not a part of the height of the triangle . I will call it D
- DL²=(KL/2)²+D²
- 52/4= 9+ D²
- D² = 52/4-9 +4 SO D=2
now the height of the trigle is H= BD-D= 8-2=6
NOw the area of the triangle is :
- A=(KL*H)/2 ⇒ A= (6*6)/2=18
THE ANSWER IS 18 SQ.UN