Tossing a die will have 6 possible outcomes. Those are having sides that are number 1 to 6. The sample space of tossing 3 dice is equal to 6³ which is equal to 216. Now for the calculation of probabilities,
P(two 5s) = (1 x 1 x 5)/216
As we have to have the 5 in the die for two times, then for the 1 time, we can have all other numbers except 5. The answer is 5/216.
P(three 5s) = (1 x 1 x 1)/216 = 1/216
P(one 5 or two 5s) = (1 x 5 x 5)/216 + (1 x 1 x 5)/216 = 5/36
Answer:
79.5
Step-by-step explanation:
Answer:
Step-by-step explanation:
First confirm that x = 1 is one of the zeros.
f(1) = 2(1)^3 - 14(1)^2 + 38(1) - 26
f(1) = 2 - 14 + 38 - 26
f(1) = -12 + 38 = + 26
f(1) = 26 - 26
f(1) = 0
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next perform a long division
x -1 || 2x^3 - 14x^2 + 38x - 26 || 2x^2 - 12x + 26
2x^3 - 2x^2
===========
-12x^2 + 28x
-12x^2 +12x
==========
26x -26
26x - 26
========
0
Now you can factor 2x^2 - 12x + 26
2(x^2 - 6x + 13)
The discriminate of the quadratic is negative. (36 - 4*1*13) = - 16
So you are going to get a complex result.
x = -(-6) +/- sqrt(-16)
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2
x = 3 +/- 2i
f(x) = 2*(x - 1)*(x - 3 + 2i)*(x - 3 - 2i)
The zeros are
1
3 +/- 2i
Answer:
The -7 changes to 7 and the ordered pair of the new point is (4,7)
The -2 changes to 2 and the ordered pair of the new point is (2,-5)
Step-by-step explanation:
Hope this helped.
