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3 years ago

What is the correct prime factorization of 24?

Mathematics

2 answers:

Sav [38]3 years ago

8 0

Answer:

24= 2*2*2*3= 2^3*3

Step-by-step explanation:

Dmitry [639]3 years ago

6 0

Answer:

prime factorization of 24 = 2 * 2 * 2 * 3 = 2^3 * 3

Step-by-step explanation:

List the first several prime numbers: 2, 3, 5, 7, 11, 13, ...

Now start with 2. Divide 24 by 2 as many times as possible. Then move on to 3, to 5, etc. until you end up with 1.

Divide only by prime numbers.

Start by dividing 24 by 2:

24/2 = 12 12 is divisible by 2, so divide 12 by 2.

12/2 = 6 6 is also divisible by 2, so divide 6 by 2.

6/2 = 3 Since 3 is not divisible by 2, move onto 3.

3/3 = 1 When you get to 1, stop. The prime factors are all the prime numbers in boldface that you divided by.

prime factorization of 24 = 2 * 2 * 2 * 3 = 2^3 * 3

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3 years ago

Read 2 more answers

A pen company averages 1.2 defective pens per carton produced (200 pens). The number of defects per carton is Poisson distribute

nlexa [21]

Answer:

a. P(x = 0 | λ = 1.2) = 0.301

b. P(x ≥ 8 | λ = 1.2) = 0.000

c. P(x > 5 | λ = 1.2) = 0.002

Step-by-step explanation:

If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:

$P(k)=\frac{\lambda^{k}e^{-\lambda}}{k!}= \frac{1.2^{k}\cdot e^{-1.2}}{k!}$

a. What is the probability of selecting a carton and finding no defective pens?

This happens for k=0, so the probability is:

$P(0)=\frac{1.2^{0}\cdot e^{-1.2}}{0!}=e^{-1.2}=0.301$

b. What is the probability of finding eight or more defective pens in a carton?

This can be calculated as one minus the probablity of having 7 or less defective pens.

$P(k\geq8)=1-P(k$

$P(0)=1.2^{0} \cdot e^{-1.2}/0!=1*0.3012/1=0.301\\\\P(1)=1.2^{1} \cdot e^{-1.2}/1!=1*0.3012/1=0.361\\\\P(2)=1.2^{2} \cdot e^{-1.2}/2!=1*0.3012/2=0.217\\\\P(3)=1.2^{3} \cdot e^{-1.2}/3!=2*0.3012/6=0.087\\\\P(4)=1.2^{4} \cdot e^{-1.2}/4!=2*0.3012/24=0.026\\\\P(5)=1.2^{5} \cdot e^{-1.2}/5!=2*0.3012/120=0.006\\\\P(6)=1.2^{6} \cdot e^{-1.2}/6!=3*0.3012/720=0.001\\\\P(7)=1.2^{7} \cdot e^{-1.2}/7!=4*0.3012/5040=0\\\\$

$P(k$

c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?

We can calculate this as we did the previous question, but for k=5.

$P(k>5)=1-P(k\leq5)=1-\sum_{k=0}^5P(k)\\\\P(k>5)=1-(0.301+0.361+0.217+0.087+0.026+0.006)\\\\P(k>5)=1-0.998=0.002$

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Answer:

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Step-by-step explanation:

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3 years ago

What is the correct prime factorization of 24?​

What is the correct prime factorization of 24?