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GaryK [48]
3 years ago
15

Um condutor acumulou 13 pontos em infração determine todas as possibilidades

Mathematics
1 answer:
777dan777 [17]3 years ago
3 0
Oysyoydoyfoyxouxougdfyyiyydouf
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(–5.7 • 2) • 8 = –5.7 • (2 • x)
pashok25 [27]
It's 8 because of the commutative property.
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PLEASE PLEASE PLEASE HELP ME WITH ALGEBRA 1 PLEASE!?!?!?!?!!!?!?!?!?!?! I WILL GIVE LOTS OF POINTS AND BRAINLEST PLEEEEEEEEEEEEE
jekas [21]

Answer:

  • there are no solutions (lines do not intersect)
  • there is one solution (lines intersect at one point)
  • there are an infinite number of solutions (lines overlap—are the same line)

Step-by-step explanation:

"A system of linear equations" covers a lot of territory. In Algebra 1, it usually means two linear equations in two unknowns. Each of those equations will graph as a line on a coordinate plane.

A solution is a point that satisfies all the equations. That is, it is a point that is on all the lines described by the system of equations.

The geometry of lines on a plane comes into play with regard to solutions.

  • The lines may be parallel, hence never intersect. (<em>No points</em> will be on all the lines.)
  • The lines may intersect at <em>one point</em>.
  • The lines may be the same line, overlapping, identical, coincident, consisting of <em>all the same points, an infinite number</em>.

4 0
3 years ago
Find the quotient in their exponential form.
vodomira [7]

Answer:

0.609663161

Step-by-step explanation:

(8/27)power5 = 5.57533755

(2/3)power7= 9.14494717

then, u just devide... from what i know

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3 years ago
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Can 18/33 be simplified
oksian1 [2.3K]
Yes  because both 18 and 33 are mulyiples of 3

18/33  = 6 / 11
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3 years ago
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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
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