20-x2/y-x for x=2 and y=6

2 answers:

8 0

$\text{Hello there!}\\\\\text{Plug in the numbers and solve:}\\\\\frac{20-x^2}{y-x}\\\\\frac{20-2^2}{6-2}\\\\\frac{20-4}{6-2}\\\\\frac{16}{4}\\\\=4\\\\\boxed{\text{Answer: 4}}$

MaRussiya [10]3 years ago

7 0

Use substitution to solve the equation:
20-4/4.
4/4=1 so it simplifies to 20-1 which equals 19

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In a multiple choice quiz there are 5 questions and 4 choices for each question (a, b, c, d). Robin has not studied for the quiz

Ahat [919]

Answer:

a) There is a 18.75% probability that the first question that she gets right is the second question.

b) There is a 65.92% probability that she gets exactly 1 or exactly 2 questions right.

c) There is a 10.35% probability that she gets the majority of the questions right.

Step-by-step explanation:

Each question can have two outcomes. Either it is right, or it is wrong. So, for b) and c), we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

In this problem we have that:

Each question has 4 choices. So for each question, Robin has a $\frac{1}{4} = 0.25$ probability of getting ir right. So $\pi = 0.25$ . There are five questions, so $n = 5$ .

(a) What is the probability that the first question she gets right is the second question?

There is a 75% probability of getting the first question wrong and there is a 25% probability of getting the second question right. These probabilities are independent.

$P = 0.75(0.25) = 0.1875$