How many solutions does the nonlinear system of equations graphed below have?

Mathematics

2 answers:

zhannawk [14.2K]3 years ago

8 0

The answer is Three - apex !!!

NeTakaya3 years ago

3 0

Answer:

Three solutions.

Step-by-step explanation:

In order to solve system of equations by graphical method, we graph the system of equations. The intersection point (s) of the two graph would give us the solution.

If the graph of the equations intersect at 2 points then there will be 2 solutions, if there are 4 intersection then there would be 4 solutions and so on.

For the given graph, we can see that there are three intersection points.

Hence, there would be three solutions.

You might be interested in

According to the National Bridge Inspection Standard (NBIS), public bridges over 20 feet in length must be inspected and rated e

slamgirl [31]

Answer:

1.80% probability that in a random sample of 12 major Denver bridges, at least 4 will have an inspection rating of 4 or below in 2020.

Step-by-step explanation:

For each bridge, there are only two possible outcomes. Either it has rating of 4 or below, or it does not. The probability of a bridge being rated 4 or below is independent from other bridges. So 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.

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

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

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

And p is the probability of X happening.

For the year 2020, the engineers forecast that 9% of all major Denver bridges will have ratings of 4 or below.

This means that $p = 0.09$

Use the forecast to find the probability that in a random sample of 12 major Denver bridges, at least 4 will have an inspection rating of 4 or below in 2020.

Either less than 4 have a rating of 4 or below, or at least 4 does. The sum of the probabilities of these events is 1.

$P(X < 4) + P(X \geq 4) = 1$