1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
aleksklad [387]
3 years ago
12

In the diagram to the right three lines intersect at n.The measure of < GNF is 60, and the measure of < MNL is 47. What is

the measure of < HNK?

Mathematics
1 answer:
Valentin [98]3 years ago
8 0
< GNL = 60
< MNL = 47

< MNL and < GNH are vertical angles and are equal
so < GNH = 47

< GNL + < GNH + < HNK = 180....these angles form a straight line...so when added, they equal 180

60 + 47 + < HNK = 180
107 + < HNK = 180
< HNK = 180 - 107
< HNK = 73 <====




You might be interested in
B=(2x+3)(4x^2-6x+9)-2(4^3-1)
nexus9112 [7]
B = (2x+3)(4x^2-6x+9)-2(4^3-1)
B = 8x^3-99

Hope it helps : )
5 0
2 years ago
Idc anymore bc i got my answer XDD
kozerog [31]

Answer: hows are you den

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Find the gradient of the line segment between points (8,6) and (10,14)
fenix001 [56]

Answer: 4

Step-by-step explanation:

3 0
2 years ago
Help me with Question 4
ruslelena [56]

Answer:

a

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
A coin, having probability p of landing heads, is continually flipped until at least one head and one tail have been flipped. (a
Natali [406]

Answer:

(a)

The probability that you stop at the fifth flip would be

                                   p^4 (1-p)  + (1-p)^4 p

(b)

The expected numbers of flips needed would be

\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1}  = 1/p

Therefore, suppose that  p = 0.5, then the expected number of flips needed would be 1/0.5  = 2.

Step-by-step explanation:

(a)

Case 1

Imagine that you throw your coin and you get only heads, then you would stop when you get the first tail. So the probability that you stop at the fifth flip would be

p^4 (1-p)

Case 2

Imagine that you throw your coin and you get only tails, then you would stop when you get the first head. So the probability that you stop at the fifth flip would be

(1-p)^4p

Therefore the probability that you stop at the fifth flip would be

                                    p^4 (1-p)  + (1-p)^4 p

(b)

The expected numbers of flips needed would be

\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1}  = 1/p

Therefore, suppose that  p = 0.5, then the expected number of flips needed would be 1/0.5  = 2.

7 0
3 years ago
Other questions:
  • Please help as I can’t figure this out :(
    10·1 answer
  • PLZ HELP RN. why do we learned about triangles before parallelograms?
    11·1 answer
  • FIRST ONE TO ANSWER GET'S A BRAINLIEST!!! PLZ HELP ME!!!
    14·2 answers
  • Thirty-five students belong to the science club. This 1/4 the number in the 5 sections of the eighth grade. How many students ar
    9·1 answer
  • PLEASE PLEASE HELP I HAVE OTHER WORK TO DO AND I NEED THIS ANSWER ASAP
    9·2 answers
  • Jason can travel 24 miles in 1/2 hour. What is his average speed in miles per hour
    13·1 answer
  • Which triangle is both scalone and acuto?
    12·1 answer
  • If f(x) = 5x2 - 3x , what is f(-3)
    6·2 answers
  • Helpppp match each of the following equations according to their slope and y-intercept
    11·1 answer
  • The total cost of 3 bikes is 480 000 . One bike costs twice as much as the smallest and another 3 times as much a the smallest ?
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!