Factor the GCF: −12x4y − 9x3y2 + 3x2y3

1 answer:

monitta3 years ago

3 0

Factor the coefficients:
-12=(-1)(3)(2^2)
-9=(-1)(3^2)
3=3

The greatest common factor (GCF) is 3

Next we find the GCF for the variable x.
x^4
x^3
x^2
The GCF is x^2.

Next GCF for variable y.
y
y^2
y^3
the GCF is y

Therefore the GCF is 3x^2y
To factor this out, we need to divide each term by the GCF,
(3x^2y)(−12x4y/(3x^2y) − 9x3y2/(3x^2y) + 3x2y3/(3x^2y) )
=(3x^2y)(-4x^2-3xy+y^2)
if we wish, we can factor further:
(3x^2y)(y-4x)(x+y)

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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

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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:

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