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2 years ago

Plz help with any problems you can, show work for number 29

Mathematics

1 answer:

alexgriva [62]2 years ago

5 0

Answer:

For number 13, I think it is rational, integer, and whole number

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I cant find answer pls help me :)

aleksandr82 [10.1K]

Answer:

m n^4 p^3

Step-by-step explanation:

m^7 n^4 p^3 and m n^12 p^5

We have 7 m's on the left and one on the right

We have one m in common

We have 4 n's on the left and 12 on the right

We have 4 n's in common

We have 3 p's on the left and 5 on the right

We have 3 p's in common

m n^4 p^3

3 0

3 years ago

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HELPPPPP !!!!! Answers ????

tamaranim1 [39]

All 3 sides are all equal

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3 years ago

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(7-13x^3-11x)-(2x^3+8-4x^5)

riadik2000 [5.3K]

Answer:

4x^5−15x^3−11x−1

Step-by-step explanation:

Simplify the expression. 4x^5−15x^3−11x−1

5 0

3 years ago

An automated egg carton loader has a 1% probability of cracking an egg, and a customer will complain if more than one egg per do

vaieri [72.5K]

Answer:

a) Binomial distribution B(n=12,p=0.01)

b) P=0.007

c) P=0.999924

d) P=0.366

Step-by-step explanation:

a) The distribution of cracked eggs per dozen should be a binomial distribution B(12,0.01), as it can model 12 independent events.

b) To calculate the probability of having a carton of dozen eggs with more than one cracked egg, we will first calculate the probabilities of having zero or one cracked egg.

$P(k=0)=\binom{12}{0}p^0(1-p)^{12}=1*1*0.99^{12}=1*0.886=0.886\\\\P(k=1)=\binom{12}{1}p^1(1-p)^{11}=12*0.01*0.99^{11}=12*0.01*0.895=0.107$

Then,

$P(k>1)=1-(P(k=0)+P(k=1))=1-(0.886+0.107)=1-0.993=0.007$

c) In this case, the distribution is B(1200,0.01)

$P(k=0)=\binom{1200}{0}p^0(1-p)^{12}=1*1*0.99^{1200}=1* 0.000006 = 0.000006 \\\\ P(k=1)=\binom{1200}{1}p^1(1-p)^{1199}=1200*0.01*0.99^{1199}=1200*0.01* 0.000006 \\\\P(k=1)= 0.00007\\\\\\P(k\leq1)=0.000006+0.000070=0.000076\\\\\\P(k>1)=1-P(k\leq 1)=1-0.000076=0.999924$

d) In this case, the distribution is B(100,0.01)

We can calculate this probability as the probability of having 0 cracked eggs in a batch of 100 eggs.

$P(k=0)=\binom{100}{0}p^0(1-p)^{100}=0.99^{100}=0.366$

5 0

3 years ago

PLZZ NEED HELP ASAPPP :((

kherson [118]

Answer: I would say it hundredth place

Step-by-step explanation:

3 0

2 years ago

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