Using a number cube and this hat with 5 different-colored marbles, which simulation would help you answer this question? During
the first 70 days of school, Becca’s science class does a lab experiment 2/3 of the time. The class meets Monday through Friday. If you choose a class at random, what are the chances that Becca did a lab experiment on a Friday?
1 answer:
Answer:
see the attachment
Step-by-step explanation:
We assume that the question is interested in the probability that a randomly chosen class is a Friday class with a lab experiment (2/15). That is somewhat different from the probability that a lab experiment is conducted on a Friday (2/3).
Based on our assumption, we want to create a simulation that includes a 1/5 chance of the day being a Friday, along with a 2/3 chance that the class has a lab experiment on whatever day it is.
That simulation can consist of choosing 1 of 5 differently-colored marbles, and rolling a 6-sided die with 2/3 of the numbers being designated as representing a lab-experiment day. (The marble must be replaced and the marbles stirred for the next trial.) For our purpose, we can designate the yellow marble as "Friday", and numbers greater than 2 as "lab-experiment".
The simulation of 70 different choices of a random class is shown in the attachment.
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<em>Comment on the question</em>
IMO, the use of <em>70 trials</em> is coincidentally the same number as the first <em>70 days</em> of school. The calendar is deterministic, so there will be exactly 14 Fridays in that period. If, in 70 draws, you get 16 yellow marbles, you cannot say, "the probability of a Friday is 16/70." You need to be very careful to properly state the question you're trying to answer.
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Answer:
Solving the given linear system, we get x = 1 and y = -3.
The solution set is: (1,-3)
Step-by-step explanation:
We need to solve the linear system of equations.
We can write second equation y=x-4 as:
Let:
Now, Adding equation 1 and 2
So, we get x = 1
Now, put value of x in second equation to find value of y:
So, we get y = -3
Solving the given linear system, we get x = 1 and y = -3.
The solution set is: (1,-3)
Answer: The correct options are (1) (5,10), (2) (3,-3), (3) x = -1, (4) , (5) 21s and (6) 0, -1, and 5.
Explanation:
Te standard form of the parabola is,
.....(1)
Where, (h,k) is the vertex of the parabola.
(1)
The given equation is,
Comparing this equation with equation (1),we get,
and
Therefore, the vertex of the graph is (5,10) and the fourth option is correct.
(2)
The given equation is,
To make perfect square add , i.e.,
. Since there is factor 3 outside the parentheses, so subtract three times of 9.
Comparing this equation with equation (1),we get,
and
Therefore, the vertex of the graph is (3,-3) and the fourth option is correct.
(3)
The given equation is
To make perfect square add , i.e.,
. Since there is factor 4 outside the parentheses, so subtract three times of 1.
Comparing this equation with equation (1),we get,
and
The vertex of the equation is (-1,3) so the axis is x=-1 and the correct option is 2.
(4)
The given equation is,
To make perfect square add , i.e.,
.
Therefore, the correct option is 4.
(5)
The given equation is,
It can be written as,
It is a downward parabola. so the maximum height of the function is on its vertex.
The x coordinate of the vertex is,
Therefore, after 21 seconds the projectile reach its maximum height and the correct option is first.
(6)
The given equation is,
Use factoring method to find the factors of the equation.
Equate each factor equal to 0.
Therefore, the zeros of the given equation is 0, -1, 5 and the correct option is 2.