HELP ME IF RIGHT ILL MARK YOU BRAINLIEST
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Answer:
12.08
Step-by-step explanation:
For each sunflower, there are only two possible outcomes. Either it germinates, or it does not. The probability of a sunflower germinating is independent of other sunflowers. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
The standard deviation of the binomial distribution is:
Seventy-six percent of sunflower seeds will germinate into a flower
This means that
Samples of 800:
This means that
The standard deviation for the number of sunflower seeds that will germinate in such samples of size 800 is:
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