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Ahat [919]
4 years ago
14

Round 939,515 to the nearest ten.

Mathematics
1 answer:
Effectus [21]4 years ago
4 0

Answer: 939,520

Step-by-step explanation: To round 939,515 to the nearest ten, we first find the digit in the rounding place which in this case is the 1 in tens place.

Next, we look at the digit to the right of of the 1 which in this case is 5.

According to the rules of rounding, since the digit to the right of the rounding place is greater than or equal to 5, we round up.

This means that we add 1 to the digit that's in the rounding place so 1 will become 2 and we change all the digits that appear to the right of the rounding place to 0.

So 939,515 rounded to the nearest 10 is 939,520.

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A coin is tossed 10 times. given that the first 9 tosses were heads, find the percent chance of getting 10 heads in a row?
Olin [163]

Answer:

1/2

Step-by-step explanation:

Assume the coin is a fair coin.

Since we already know that the first 9 tosses were heads, the probability of 9 straight heads in the first 9 tosses is 1.

It all depends on the last toss. There is an equal probability of heads and tails in a single toss of a fair coin.

p(heads) = 1/2

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Answer:

v = 25

Step-by-step explanation:

The crucial information you need to know to solve this is to realize that HI and GH are the same length. However, why they are equal is not immediately obvious.

Both sides of the middle line (HF) are symmetrical, since G and I are the same distance away from the line, and they both lie on a line perpendicular to the middle line.

Note: we know they're the same distance away due to the small red marks in the lines, indicating that they're the same length.

The angles at G and I in the triangles are also the same, as the lines from G and I both meet at H. If they were different angles, they would each hit a different point on the middle line.

Thus, we can conclude that GH and HI are the same length.

Since we know the following:

GH = 4v - 75

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We can set GH and HI equal to each other and solve the equation.

4v - 75 = v

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\omega (1)=2\pi\\\\\omega=2\pi

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