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nekit [7.7K]
3 years ago
6

The length of a hyperbolas transverse axis is equal to the what the distances from any point on the hyperbola to each focus

Mathematics
2 answers:
konstantin123 [22]3 years ago
8 0

Answer:

yes i agree i took the test

Step-by-step explanation:


ANEK [815]3 years ago
5 0
Difference between -apex
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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
Which sequence of rigid transformations will map the preimage ABC onto image A'B'C' ?
iren2701 [21]

Answer:

The answer is option A, a double reflection

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Help? 10 points please ❤
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Total Surface Area = 2(8.2 x 3.4) + 2(8.2 x 2) + 2(3.4 x 2)
Total Surface Area = 36.56 in²

Question 25

Total Surface Area = 2(10 x 12) + 2(12 x 4) + 2(10 x 4)
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4 years ago
Help meeeeeeee!!! REEEE
Alina [70]

Answer:

I think 5 faces and 10 edges but im not sure what a vertice is sorry

Step-by-step explanation:

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3 years ago
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