If t the possibilities the to win are 1/16,.then the possibilities to lose are 15/16
16/16-1/16=15/16
so every 15 of the 16 times they will raise 25 cents which is almost 100% of the times
Every15 of 16 times they will raise $15.But also they need to give away$5 price
so $10 every 15/16
Guys Here Is Your Answer
-3x-3<6
-3x<6+3
(collection of like terms).
-3x<9
-3x>9
( the sign changes because of the negative sign -).
-3x/-3>9/-3
(divide by the coefficient of x)
x>-3
for the graph
see this
Where’s the graph? I need to see the graph
You might need to consult your teacher or text to learn the details of the "ratio of perfect squares method" of determining an approximate square root. No reference to such a method can be found in an Internet search except in conjunction with problems similar to this one.
A method that can be used to find a first approximation of a square root is linear interpolation between the roots of adjacent perfect squares. For this, ...
• find the perfect squares of the consecutive integers that lie on either side of the root of interest
• form the ratio of difference between the number and the smaller square and the difference between the squares
• add this ratio to the smaller of the two integers to get an approximation of the root.
In this case, the square root of 96 lies between 9² = 81 and 10² = 100. The ratio of interest is (96 -81)/(100 -81) = 15/19. The approximate square root of 96 is then ...
√96 ≈ 9 + 15/19 ≈ 9.8
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If a = floor(√n), then this approximation to the root can be written as
√n ≈ a + (n -a²)/(2a+1)
If we define b = n - a², this looks like √n ≈ a + b/(2a+1). The approximation can be refined by replacing the 1 in the denominator with (b/(2a+1)). Repeately doing this replacement results in a "continued fraction" that converges to √n as more layers are added.
Example:
123456 in scientific notation would be
1.23456x10^5
you rewrite the number so it is less than 10 but more than 1 and multiply it by 10 to the power of however many decimal places it takes to get the number back to its original form
