Jeff has a box containing 26 tiles, each with a different letter of the alphabet. he randomly draws 4 letters, one at a time wit
1 answer:
The probability to choose letter n is 
. After drawing letter n there left 25 letters and the probability to choose the letter w is 
. After drawing two letters n and w there left 24 letters and then the probability to choose letter b is 
. After drawing three letters n, w and b there left 23 letters, so the probability to choose letter t is 
.
Using the product rule for probabilities, you can obtain that
.
Using the product rule for probabilities, you can obtain that
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QUESTION 1
The given system of equations is
To solve by linear combination, we add equation (1) to equation (2) to get,
We divide through by 4 to obtain,
We put d=3 into equation (2) to get,
QUESTION 2
The given system is
4x + y = 5 ...eqn(1)
3x + y = 3 ...eqn(2)
To solve by linear combination, we subtract equation (2) from equation (1) to eliminate y from the equation.
This will give us,
This implies that,
Put x=3 into equation (1) to get,
The solution is
QUESTION 3
We want to solve the system;
a – 2b = –2 ....eqn(1)
2a + 2b = 14...eqn(2)
by linear combination.
We need to add equation (1) to equation (2) to eliminate b.
This implies that,
Simplify,
Divide both sides by 3 to get,
Put a=4 into equation (2) to obtain,
The ordered pair in the form (a, b) is
QUESTION 4
The given system of equations is
11x + 4y = 18 ...eqn(1)
3x + 4y = 2 ...eqn(2)
We subtract equation (2) from equation (1) to get,
Put x=2 into equation (2) to obtain,
This implies that,
The correct answer is (2,-1).
QUESTION 5
The given system is ;
2d + e = 8...eqn1
d – e = 4...eqn2
We add the two equations to eliminate e.
This implies that,
We divide both sides by 3 to get,
We put d=4 into equation (2) to get,
The solution is
The given system of equations is
To solve by linear combination, we add equation (1) to equation (2) to get,
We divide through by 4 to obtain,
We put d=3 into equation (2) to get,
QUESTION 2
The given system is
4x + y = 5 ...eqn(1)
3x + y = 3 ...eqn(2)
To solve by linear combination, we subtract equation (2) from equation (1) to eliminate y from the equation.
This will give us,
This implies that,
Put x=3 into equation (1) to get,
The solution is
QUESTION 3
We want to solve the system;
a – 2b = –2 ....eqn(1)
2a + 2b = 14...eqn(2)
by linear combination.
We need to add equation (1) to equation (2) to eliminate b.
This implies that,
Simplify,
Divide both sides by 3 to get,
Put a=4 into equation (2) to obtain,
The ordered pair in the form (a, b) is
QUESTION 4
The given system of equations is
11x + 4y = 18 ...eqn(1)
3x + 4y = 2 ...eqn(2)
We subtract equation (2) from equation (1) to get,
Put x=2 into equation (2) to obtain,
This implies that,
The correct answer is (2,-1).
QUESTION 5
The given system is ;
2d + e = 8...eqn1
d – e = 4...eqn2
We add the two equations to eliminate e.
This implies that,
We divide both sides by 3 to get,
We put d=4 into equation (2) to get,
The solution is