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lapo4ka [179]
2 years ago
7

A bag contains 4 blue and 5 red counters. Two counters are chosen at random without replacement. What is the probability that th

e counters are both red?
Mathematics
1 answer:
jeka942 years ago
6 0

Answer:

P(both are red) = 5/18

Step-by-step explanation:

total counters = 9

red counters = 5

P(both red) = P(R1) n P(R2)

5/9 * 4/8 =5/18

5/9 because the red counters are 5 and the total number of counters are 9.

I multiplied by 4/8 because, after selecting the first red counter without replacement, the remaining is 4 and the total decreases to 8.

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3 0
3 years ago
Squareroot(y-1) + 3 = y
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Note:xy means x times y and x(y) means the same thing

so
first we get rid of square root then
make the equation equal to zero becaues if
xy=0 then x or/and y=0



squareroot(y-1)+3=y

isolate the squareroot
subtrac 3 from boht sides

squareroot(y-1)=y-3

square both sides (since they are equal, you should be able to square both sides and still make it true)

(squareroot(y-1))^2=(y-3)^2
(y-1)=(y-3)(y-3)
y-1=y^2-6y+9
subtrac y from both sides
-1=y^2-7y+9
add 1 to both sides
0=y^2-7y+10
find what two number multiply to make 10 and add to get -7
the answer is -2 and -5

0=(y-5)(y-2)
therfore
y-5=0
and/or
y-2=0

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let's try out 2
square root(2-1)+3=2
square root(1)+3=2
1+3=2
false
so 2 doesn't work



let's try 5
squareroot(5-1)+3=5
squareroot(4)+3=5
2+3=5
5=5
true

y=5
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3 years ago
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