Answer:
0.1686 = 16.86% probability that a student will answer exactly 6 questions correct if he makes random guesses on all 20 questions.
Step-by-step explanation:
For each question, there are only two possible outcomes. Either the students answer it correctly, or he/she does not. The probability of answering a question correctly is independent of any other question. This means that the binomial probability distribution is used to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which 
is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
20 questions:
This means that 
.
Answer choices are A, B, C and D:
Only one is correct, so 
Find the probability that a student will answer exactly 6 questions correct if he makes random guesses on all 20 questions.
This is P(X = 6). So
0.1686 = 16.86% probability that a student will answer exactly 6 questions correct if he makes random guesses on all 20 questions.
Answer:
3x-y= -8
Step-by-step explanation:
i believe this answer is correct
Answer:
57
Step-by-step explanation:
Answer:
B
Step-by-step explanation:
The answer is B because if you move two times to the right it gives you 0.09 like the eqution.
But scientific notation requires number without decimal, so the answer cannot be E because 0.9 is a decimal.
Just reminder: B and D have the same result!! just different coefficient
Answer:
a) (1, 2) not on the graph.
b) (1, -1) is on the graph.
Step-by-step explanation:
Given the equation of the line as:
The points given are
a) (1, 2)
To determine whether it is on the graph of the line or not.
To do so, we can do 2 things:
1. Draw the graph and plot the point on the graph to check whether it is on the graph or not.
2. To put the point in the given equation of the line, whether the equation is satisfied or not.
For method 1: Kindly refer to the attached image of the line and point plotted.
Method 2:
Let us put 
in the Left Hand Side (LHS) of equation.
(1, 2) Not on the graph.
(b) (1, -1)
For method 1: Kindly refer to the attached image of the line and point plotted.
Method 2:
Let us put 
in the Left Hand Side (LHS) of equation.
(1, -1) is on the graph.
