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goblinko [34]
2 years ago
10

Line k is the perpendicular bisector of pq and point r which of the following statements must be true?

Mathematics
1 answer:
Artist 52 [7]2 years ago
3 0

Answer:

question incomplete.

Step-by-step explanation:

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PLZ HELPPPPPPPPPPPPPPPPPPPPPPPP
-BARSIC- [3]

Answer:

I think the awnser is probably b

3 0
3 years ago
Read 2 more answers
A taxi charges 2.50 plus .50 fee for each mile. If the ride costs 7.50 how many miles did he travel
NeX [460]

Answer:<em> 10 miles</em>

Step-by-step explanation:

<em>Mile 1: 3.00</em>

<em>Mile 2: 3.50</em>

<em>Mile 3: 4.00</em>

<em>Mile 4: 4.50</em>

<em>Mile 5: 5.00</em>

<em>Mile 6: 5.50</em>

<em>Mile 7: 6.00</em>

<em>Mile 8: 6.50</em>

<em>Mile 9: 7.00</em>

<em>Mile 10: 7.50</em>

7 0
2 years ago
The diameter of a wheel of a car is 0.75 m. if the car travels at an average speed of 14 m/s, find the
raketka [301]

Step-by-step explanation:

Circumference of wheel = πd = 0.75πm.

How far the car travels per minute

= 60s * (14m/s) = 840m.

Hence number of revolutions per minute

= 840m / 0.75πm

= 356.507....

= 357. (nearest whole number)

6 0
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
Side :44 mm,28 mm,24 mm what is the triangle
bagirrra123 [75]
This is an scalene triangle because a scalene triangle has no equal size
4 0
3 years ago
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