Hey there!!
Given equation :
... -4 ( x + 10 ) - 6 = -3 ( x - 2 )
Using the distributive property :
... -4x - 40 - 6 = -3x + 6
Combining like terms :
... -4x - 46 = -3x + 6
Adding 46 on both sides :
... -4x = -3x + 52
Adding 3x on both sides :
... -x = 52
... x = -52
Hope helps!
Answer:
-11 for the first one and 0 for the second one
Step-by-step explanation:
Answer:
0.430625=0.431
Step-by-step explanation:
Answer:
0.430625 = 0.431
Step-by-step explanation:
Let W represent winning, D represent a draw and L represent a loss.
12+ points can be garnered in each of the following ways.
6W 0D 0L
5W 1D 0L
5W 0D 1L
4W 2D 0L
4W 1D 1L
4W 0D 2L
3W 3D 0L
The probability of getting 12+ points is the sum of all these 7 probabilities.
Knowing that P(W) = 0.5
P(D) = 0.1
P(L) = 0.4
P(6W 0D 0L) = [6!/(6!0!0!)] 0.5⁶ 0.1⁰ 0.4⁰ = 0.015625
P(5W 1D 0L) = [6!/(5!1!0!)] 0.5⁵ 0.1¹ 0.4⁰ = 0.01875
P(5W 0D 1L) = [6!/(5!0!1!)] 0.5⁵ 0.1⁰ 0.4¹ = 0.075
P(4W 2D 0L) = [6!/(4!2!0!)] 0.5⁴ 0.1² 0.4⁰ = 0.09375
P(4W 1D 1L) = [6!/(4!1!1!)] 0.5⁴ 0.1¹ 0.4¹ = 0.075
P(4W 0D 2L) = [6!/(4!0!2!)] 0.5⁴ 0.1⁰ 0.4² = 0.15
P(3W 3D 0L) = [6!/(3!3!0!)] 0.5³ 0.1³ 0.4⁰ = 0.0025
The probability of getting 12+ points = 0.015625 + 0.01875 + 0.075 + 0.09375 + 0.075 + 0.15 + 0.0025 = 0.430625
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9514 1404 393
Answer:
x = 3/2
Step-by-step explanation:
Multiply by 3 and solve the resulting 2-step equation.
5 = (1/3)(2x +12)
15 = 2x +12 . . . . . . multiply by 3
3 = 2x . . . . . . . . . . subtract 12
3/2 = x . . . . . . . . . . divide by 2
Answer:
A
Step-by-step explanation:
<em>Okay so, im not the best at explaining, but so we have a percentage. its 25%. you want to take the 25% and put it over 100 because a percentage is a number or ratio expressed as a fraction of 100. so it will be written as 25%/100</em>
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<em>divide 25 by 100 and you get 0.25, correct? so that's one step further towards the answer.</em>
<em>then we write x/90 because we don't know (well we do) what we're gonna put. but, we're gonna multiply 0.25 by 90 and we get 22.5, which is equivalent to </em><em>22.50</em><em>.</em>