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kifflom [539]
3 years ago
14

**hint** Find the hypotenuse and then subtract that from the distance he actually traveled.

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

Answer:

10 miles

Step-by-step explanation:

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Identify the Quadrant where the point is located.
ValentinkaMS [17]

Answer:

1 ko second quadrant and 2 ko chai fourth quadrant

Step-by-step explanation:

plz mark me as brainest answer

8 0
3 years ago
WILL MARK BRAINLIEST IF GOTTEN RIGHT
kkurt [141]

Answer:

85

Step-by-step explanation:

Dont need to explain it quik mafs

6 0
3 years ago
Please i’m not good at math i need help i’m doing summer school online and i need help
Sati [7]

Answer:

135

____________

2*25/5=10

100*5=500

15*5=75

75*5=375

10+500-375=135

Hope this helps :3

8 0
3 years ago
If xt = 3y -1 and us = 28, find the value of y
olga55 [171]
3y - 1 = 28
+ 1 + 1
----------------
3y = 29
---- ----
3 3
y= 9.66
8 0
3 years ago
Can you help me with number 6? <br> Confused abit <br> Please
Sunny_sXe [5.5K]

You can see the three diagram attached. Each link is labeled with the probability: you have probability 1/6 that a six is rolled, and 5/6 that it is not rolled.


To answer the questions, find the path that brings you to the desired outcome, and multiply all the labels you meet.


First question:

To get three sixes, you have to choose the left path at each roll. The probability is always 1/6, so the answer is


\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{6^3}


Second question:

To get no sixes, you have to choose the right path at each roll. The probability is always 5/6, so the answer is


\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{5^3}{6^3}


Third question:

To get exactly one six, it can either be the first, second or third roll.


In all cases, you have to choose the left path once and the right path twice: left-right-right mean that you get the six in the first roll, right-left-right means that you get the six in the second roll, right-right-left means that you get the six in the third roll.


In every case, the left turn has probability 1/6, and the right turn has probability 5/6. The probability of each combination is thus


\frac{1}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{5^2}{6^3}


And since there are three of these combinations, The answer is


3\frac{5^2}{6^3}


Fourth question:

Since the question suggests to use what we already achieved, let's do it: having at least one six is the complementary event of having no sixes at all. If an event has probability p, its complementary has probability 1-p. So, since the probability of no sixes is known, the probability of at least one six is


1 - \frac{5^3}{6^3}

4 0
3 years ago
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