The highest common factor of 60 and 150 is 30
And the lowest is 10
X= -5
Explaination: you subtract 12 from both sides and left with -5
Here is a system of linear equations that represents the situation.
x +5y +10z +20w = 133 . . . total amount earned
x +y +z +w = 39 . . . . . . . . . total number of bills
y = 4z . . . . . . . . . . . . . . . . . . the number of 5s is 4 times the number of 10s
x = 2y -1 . . . . . . . . . . . . . . . . the number of 1s is 1 less than twice the number of 5s
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We can substitute for x and z in the first two equations:
... (2y-1) +5y +10(y/4) +20w = 133
... (2y-1) +y +(y/4) +w = 39
These simplify to
... 9.5y +20w = 134
... 3.25y +w = 40
Solving by your favorite method, you get
... y = 12
... w = 1
So the other values can be found to be
... x = 2·12 -1 = 23
... z = 12/4 = 3
The solution to the system is (x, y, z, w) = (23, 12, 3, 1).
49. From 3 coin tosses, there are 8 possible outcomes:
... TTT, TTH, THT, THH, HTT, HTH, HHT, HHH
All but the first have at least one head, so 7/8 of the possibilities have at least one head. That's 87.5% (not among your choices).
Likewise, all but the last listed outcome have at least one tail. The problem is symmetrical that way when the coin is fair. 87.5% of outcomes have at least one tail.
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Perhaps you can tell I read your question as having two parts. If your question is the probability of getting at least one head AND at least one tail, you can see that condition includes 6 of the 8 outcomes, or 75%, matching selection d.
50. See for yourself: the calculator says 66.82%. Your best choice is selection d.
If we are to connect both tips of the cliff to the approaching ship, we form a right triangle. The height of the cliff is the opposite side of the angle of depression. The trigonometric function that would allow us to solve for the distance is tangent.
tan 18.2° = opposite /adjacent = 426 ft/ x ft
The value of x from the equation is 1295.68 ft
Thus, the answer is the second choice.
