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mestny [16]
4 years ago
6

Which dimensions will create an unlimited number of triangles?

Mathematics
2 answers:
mrs_skeptik [129]4 years ago
8 0

Answer:

B

Step-by-step explanation:

PolarNik [594]4 years ago
7 0

the answer to this question is

angle 45

angle 45

angle 90

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PLEASE ANSWER ASAP!! it might be me just being stupid but I was given a question on sequences. I was given the sequence 2,8,14,2
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I think ur homework sheet is wrong the difference is definitely 6 

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Allyson walked 10.8 km to the school. If 1 mile = 1.6 km, about how many miles did she walk?
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According to my calculations, 1.6 goes into 10.8 a total of 6 times with a remainder of 1.1999999999999993. If you continue the long division beyond the decimal point, the result would be 6.75.

Step-by-step explanation:

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Area of a rectangle 8cm long and 4cm in height. please tell me how to do this also, I'm so confused.
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4 years ago
Determine the location of a point (x, y, z) that satisfies the given condition. xyz > 0
ollegr [7]
Xyz>0, a-location of a point
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a∈{{x<0∧y<0∧z>0}∨{x>0∧y<0∧z<0}∨{x<0∧y>0∧z<0}∨{x>0∧y>0∧z>0}}
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3 years ago
Suppose the number of insect fragments in a chocolate bar follows a Poisson process with the expected number of fragments in a 2
leonid [27]

Answer:

a)The expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b)0.6004

c)19.607

Step-by-step explanation:

Let X denotes the number of fragments in 200 gm chocolate bar with expected number of fragments 10.2

X ~ Poisson(A) where \lambda = \frac{10.2}{200} = 0.051

a)We are supposed to find the expected number of insect fragments in 1/4 of a 200-gram chocolate bar

\frac{1}{4} \times 200 = 50

50 grams of bar contains expected fragments = \lambda x = 0.051 \times 50=2.55

So, the expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b) Now we are supposed to find the probability that you have to eat more than 10 grams of chocolate bar before ending your first fragment

Let X denotes the number of grams to be eaten before another fragment is detected.

P(X>10)= e^{-\lambda \times x}= e^{-0.051 \times 10}= e^{-0.51}=0.6004

c)The expected number of grams to be eaten before encountering the first fragments :

E(X)=\frac{1}{\lambda}=\frac{1}{0.051}=19.607 grams

7 0
3 years ago
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