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3 years ago

Giả thiết X là biến ngẫu nhiên X~N(µ, σ^2) thì P( |X-µ| >3,05*σ ) sẽ là

Mathematics

1 answer:

svp [43]3 years ago

7 0

Answer:

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Step-by-step explanation:

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Nico was practicing the long jump. If he jumped 4,530 centimeters, how many meters did he jump?

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45.3 meters (100 centimeters is one meter)

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2 years ago

What is the least common multiple of 15a² and 3a2? 1.3a2 2.3a3 3.15a2 4.15a3

AveGali [126]

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1. 3a2

Step-by-step explanation:

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2 years ago

Brainliest if you show how to do the work

solong [7]

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26.376 because the circumference formula is 2*3.14*radius so the radius of the goal would be 9 because radius is half of diameter which is 18 so 2*3.14*9=56.52 and the basketball's radius would be 4.8 and then 2*3.14*4.8 is 30.144 so the answer is 26.376 hope it helps

Step-by-step explanation:

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1 year ago

Read 2 more answers

How many significant digits should the product of 11,435 m and 19.35 m have

DochEvi [55]

Answer:

2 significants choes the least significant digets which is 19.35

result is 221,267.25

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2 years ago

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.

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

5 0

3 years ago