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LiRa [457]
3 years ago
14

PLEASE HELP: Find the measure of each angle in the isosceles

Mathematics
1 answer:
erma4kov [3.2K]3 years ago
3 0

Answer:

gtntt4bgt4thn

Step-by-step explanation:

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What is the value of x in x/7 = 35
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Try multiplying each side of the equation by  7 .
Then you must quickly step to one side, as the
answer jumps off the page at you.

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3 years ago
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Is this a function idk also what's the domain and range
masya89 [10]
This is a function since it passes the vertical line test. It is impossible to draw a single vertical line to have it pass through more than one point on the curve.

The domain is the set of numbers x such that x is between -3 and +3 excluding those endpoints. In other words, the domain is -3 < x < 3. We never actually get to either endpoint because of the vertical asymptotes.

The range is the set of all real numbers. It is possible to get any output we want depending on the specific input. 
3 0
4 years ago
Find the midpoint of the line segment with end coordinates of: (-2,-2) and (2,8)​
Aleonysh [2.5K]

Answer:

(0 ; 3)

Step-by-step explanation:

hello :

the midpoint of the line segment is : ((-2+2)/2 ;(-2+8)/2 )

(0 ; 3)

8 0
3 years ago
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Emma is making a scale drawing of her farm using the scale 1cm = 2.5ft. In the drawing she drew a well with a diameter of 0.5 ce
Y_Kistochka [10]
Scale : 1 cm = 2.5 ft

1 cm = 2.5 ft
0.5 cm = 2.5 x 0.5 = 1.25 ft

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Answer: 3.93 ft
8 0
3 years ago
At a specific point on a highway, vehicles arrive according to a Poisson process. Vehicles are counted in 12 second intervals, a
morpeh [17]

Answer: a) 4.6798, and b) 19.8%.

Step-by-step explanation:

Since we have given that

P(n) = \dfrac{15}{120}=0.125

As we know the poisson process, we get that

P(n)=\dfrac{(\lambda t)^n\times e^{-\lambda t}}{n!}\\\\P(n=0)=0.125=\dfrac{(\lambda \times 14)^0\times e^{-14\lambda}}{0!}\\\\0.125=e^{-14\lambda}\\\\\ln 0.125=-14\lambda\\\\-2.079=-14\lambda\\\\\lambda=\dfrac{2.079}{14}\\\\0.1485=\lambda

So, for exactly one car would be

P(n=1) is given by

=\dfrac{(0.1485\times 14)^1\times e^{-0.1485\times 14}}{1!}\\\\=0.2599

Hence, our required probability is 0.2599.

a. Approximate the number of these intervals in which exactly one car arrives

Number of these intervals in which exactly one car arrives is given by

0.2599\times 18=4.6798

We will find the traffic flow q such that

P(0)=e^{\frac{-qt}{3600}}\\\\0.125=e^{\frac{-18q}{3600}}\\\\0.125=e^{-0.005q}\\\\\ln 0.125=-0.005q\\\\-2.079=-0.005q\\\\q=\dfrac{-2.079}{-0.005}=415.88\ veh/hr

b. Estimate the percentage of time headways that will be 14 seconds or greater.

so, it becomes,

P(h\geq 14)=e^{\frac{-qt}{3600}}\\\\P(h\geq 14)=e^{\frac{-415.88\times 14}{3600}}\\\\P(h\geq 14)=0.198\\\\P(h\geq 14)=19.8\%

Hence, a) 4.6798, and b) 19.8%.

7 0
3 years ago
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