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3 years ago

I have to factor this expression (ax^2+bx+c) and I’m having some trouble. Can someone help me with this?

Mathematics

2 answers:

drek231 [11]3 years ago

6 0

3p^2 - 7p -40

Factor -7 out of -7p and rewrite -7 as -15 +8:

3p^2 +(-15 +8)p -40

Apply distributive property:

3p^2 -15p +8p -40

Group the first two terms and the last two terms:

(3p^2 -15p) + (8p-40)

Factor out the GCF from each group:

3p(p-5) + 8(p-5)

Factor the polynomial by factoring out the GCF (p-5)

(p-5)(3p+8)

Alex17521 [72]3 years ago

5 0

Ok so 3x-40=-120 and the middle number is -7 so the two numbers will have to multiply to -120 and add to - so those numbers would be 15 and-8 so then we will factor by parts so 3p^2+15p-8p-40 then section into brackets (3p^2+15p)(-8p-40) out of the first bracket you can factor out 3p and on the second bracket you can factor out -8 so you are left with 3p(p+5)-8(p+5) the combine 3p and -8 so (3p-8)(p+5)

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Step-by-step explanation:

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0.087 = 8.7% probability that this person made a day visit.

0.652 = 65.2% probability that this person made a one-night visit.

0.261 = 26.1% probability that this person made a two-night visit.

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Made a purchase.

Probability of making a purchase:

10% of 20%(day visit)

30% of 50%(one night)

20% of 30%(two night).

$p = 0.1*0.2 + 0.3*0.5 + 0.2*0.3 = 0.23$

How likely is it that this person made a day visit?

Here event B is a day visit.

10% of 20% is the percentage of purchases and day visit. So

$P(A \cap B) = 0.1*0.2 = 0.02$

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.02}{0.23} = 0.087$

0.087 = 8.7% probability that this person made a day visit.

A one-night visit?

Event B is a one night visit.

The percentage of both(one night visit and purchase) is 30% of 50%. So

$P(A \cap B) = 0.3*0.5 = 0.15$

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.15}{0.23} = 0.652$

0.652 = 65.2% probability that this person made a one-night visit.

A two-night visit?

Event B is a two night visit.

The percentage of both(two night visit and purchase) is 20% of 30%. So

$P(A \cap B) = 0.2*0.3 = 0.06$

Then

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.06}{0.23} = 0.261$

0.261 = 26.1% probability that this person made a two-night visit.

7 0

3 years ago