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sp2606 [1]
3 years ago
11

How to prove trianlges with sas sss aas

Mathematics
1 answer:
Art [367]3 years ago
7 0

Answer:

More info?

Step-by-step explanation:

If you mean how to find out each one, then

SAS: a side /, an angle <, a side /.

SSS: a side /, a side /, a side /.

AAS: an angle <, an angle <, a side /.

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Answer:

Domain: (-3,1]

Range:  [-4,0]

x-intercepts: -3, 0

y-intercepts: 0

3 0
3 years ago
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One crew can seal a parking lot in 12 hours and another in 15 hours. how long will it take to seal the parking lot if the two cr
MakcuM [25]
This is a typical work problem in algebra. The approach to this is using the equation: Rt = 1, where R is the rate per person working, t is the amount of time worked by an individual. All their rates must equal to 1. The solution is as follows:

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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

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Step-by-step explanation:

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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
Whats the answer please
lbvjy [14]
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3 years ago
What is 6,523 ÷7= what is 6523÷7=?
Nitella [24]
931.86, if rounded to the nearest hundredth. 
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