Answer:
1
1
3
3
Step-by-step explanation: just cuz
Start with 1.
1st number = 1
2nd number = 1 * 4 + 3 = 7
3rd number = 7 * 4 + 3 = 31
4th = 31 * 4 + 3 = 127
5th = 127 * 4 + 3 = 511
Which means the 6th = 511 * 4 + 3, which is 2047.
<span>The probability that one customer will place a take-out order is .45.
The probability that one customer will place an "eat in" order is .55.
Of 4 randomly selected customers, the probability that exactly 3 will order "take out" is given by the following:
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Answer:
Roots are not real
Step-by-step explanation:
To prove : The roots of x^2 +(1-k)x+k-3=0x
2
+(1−k)x+k−3=0 are real for all real values of k ?
Solution :
The roots are real when discriminant is greater than equal to zero.
i.e. b^2-4ac\geq 0b
2
−4ac≥0
The quadratic equation x^2 +(1-k)x+k-3=0x
2
+(1−k)x+k−3=0
Here, a=1, b=1-k and c=k-3
Substitute the values,
We find the discriminant,
D=(1-k)^2-4(1)(k-3)D=(1−k)
2
−4(1)(k−3)
D=1+k^2-2k-4k+12D=1+k
2
−2k−4k+12
D=k^2-6k+13D=k
2
−6k+13
D=(k-(3+2i))(k+(3+2i))D=(k−(3+2i))(k+(3+2i))
For roots to be real, D ≥ 0
But the roots are imaginary therefore the roots of the given equation are not real for any value of k.