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Sergeeva-Olga [200]
3 years ago
5

What is the value of p?

Mathematics
1 answer:
REY [17]3 years ago
5 0

Look at the picture.


The sum of measures of angles of any triangle is equal 180 °.


Therefore:


p^o+55^o+90^o=180^o\\\\p^o+145^o=180^o\ \ \ \ |-145^o\\\\p^2=35^o



Answer: C. 35°

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J is the midpoint of HK¯¯¯¯¯¯¯ . What are HJ, JK, and HK?
lana [24]
Since J is the midpoint of HK, that means HK is split into two sections HJ and JK that are the same length.

1) You are told that the m<span>easure of segment HJ = 9x-2 and that of segment JK = 4x+13. Since you also know they are equal lengths, you can set these equations equal to each other to find the value of x!
HJ = JK
</span>9x-2 = 4x+13
5x = 15
x = 3

2) Now you know x = 3. Plug that into your given equations for HJ and JK to find the length of each segment (or a shortcut would be to find one of them, and then you also know the other is the same length. I'm doing both, just to make sure I don't make a silly mistake!):
HJ = <span>9x-2 
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HJ = 27 - 2
HJ = 25

JK = 4x + 13
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JK = 12 + 13
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3) Finally, the length of HK is just the length of HJ + JK, or HK = 25 + 25 = 50.

-----

Answer: HJ = 25, JK = 25, HK = 50
4 0
3 years ago
The route used by a certain motorist in commuting to work contains two intersections with traffic signals. The probability that
tekilochka [14]

Answer:

a) P(X∩Y) = 0.2

b) P_1 = 0.16

c) P = 0.47

Step-by-step explanation:

Let's call X the event that the motorist must stop at the first signal and Y the event that the motorist must stop at the second signal.

So, P(X) = 0.36, P(Y) = 0.51 and P(X∪Y) = 0.67

Then, the probability P(X∩Y) that the motorist must stop at both signal can be calculated as:

P(X∩Y) = P(X) + P(Y) - P(X∪Y)

P(X∩Y) = 0.36 + 0.51 - 0.67

P(X∩Y) = 0.2

On the other hand, the probability P_1 that he must stop at the first signal but not at the second one can be calculated as:

P_1 = P(X) - P(X∩Y)

P_1 = 0.36 - 0.2 = 0.16

At the same way, the probability P_2 that he must stop at the second signal but not at the first one can be calculated as:

P_2 = P(Y) - P(X∩Y)

P_2 = 0.51 - 0.2 = 0.31

So, the probability that he must stop at exactly one signal is:

P = P_1+P_2\\P=0.16+0.31\\P=0.47

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