Explanation:
For the purpose of filling in the table, the BINOMPDF function is more appropriate. The table is asking for p(x)--not p(n≤x), which is what the CDF function gives you.
If you want to use the binomcdf function, the lower and upper limits should probably be the same: 0,0 or 1,1 or 2,2 and so on up to 5,5.
The binomcdf function on my TI-84 calculator only has the upper limit, so I would need to subtract the previous value to find the table entry for p(x).
Method:
First place the replace of p and r using brackets.
= 3(6) + 11.7(7)
3 times 6 is 18 and 11.7 times 7 is 81.9
= 18 + 81.9
add one and one together you get
= 99.9
<em>hope </em><em>it </em><em>helps</em>
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.
b
Step-by-step explanation: i don know
Answer:
4(x+3)
Step-by-step explanation: