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podryga [215]
3 years ago
15

Solving quadratic equations is one of the main topics of this unit. Some quadratic equations can be solved with only one method,

some can be solved with multiple methods, and some can’t be solved at all. Think about a task in your daily life that can be accomplished in a variety of ways. How do you decide the best course of action for accomplishing that task? How is this process like choosing a method for solving a quadratic equation that can be solved in multiple ways?
Mathematics
1 answer:
Mariana [72]3 years ago
7 0

Hi there!

Many things we do in everyday life have a variety of ways we can go about accomplishing them, but we most often choose the most practical and efficient method.

Efficiency saves time and prevents over-complication, which may lead to errors.

We might need to identify the specifics of the task and its circumstances to be able to determine the most efficient method to do it.

Solving a quadratic equation, we also must think about the most efficient method that can lead us to the correct answer. And doing so, we must identify the circumstances of the equation; Can it be solved by factoring? Is it easy to factor? What form is this quadratic equation in?

For example, let's say we're given the equation (x-1)(x+2)=0. This is an equation in factored form. In these kinds of scenarios, we can <em>easily</em> solve by setting each term equal to 0 (the Zero Product Property). This is the <em>most efficient </em>method:

x-1=0 --> x=1

x+2=0 --> x=-2

I hope this helps!

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at a particular high school, 42% of the students participate in sports and 25% of the students participate in drama. if 53% of t
baherus [9]

The probability that a student participates in both sports and drama is 0.14.

<h3>What is the formula for P(AUB), where A and B are any two events?</h3>

If A and B are any two events, then the probability of the joint event (A\cup B) is given by the following formula: P(A\cup B)=P(A)+P(B)-P(A\cap B)

Given that 42% of the students participate in sports and 25% of the students participate in drama and 53% of the students participate in either sports or drama.

Suppose S denotes that "a student participates in sports" and D denotes that "a student participates in drama".
So, we have P(S)=\frac{42}{100}=0.42, P(D)=\frac{25}{100}=0.25, P(S\cup D)=\frac{53}{100}=0.53.

We want to find the probability that a student participates in both sports and drama i.e., we want to find P(S\cap D).

By the above formula, we obtain:

P(S\cup D)=P(S)+P(D)-P(S\cap D)\\\Longrightarrow P(S\cap D)=P(S)+P(D)-P(S\cup D)\\\Longrightarrow P(S\cap D)=0.42+0.25-0.53\\\therefore P(S\cap D)=0.14=\frac{14}{100}

Therefore, the probability that a student participates in both sports and drama is 0.14.

To learn more about probability, refer: brainly.com/question/24756209

#SPJ9

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