4 2/3 into improper fracking would be 14/3 now times 3 , 3 and 3 go into 1 so your left with 24 so THATS the answer in other words it would be 42/3 which goes into 14 times:)
F is the anwser:))))))))))
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.
The distribution of X is X ~ N (20 , 6) and the probability that this American will receive no more than 24 Christmas cards this year is 0.7486.
<h3>Probability</h3>
a. Distribution
X ~ N (20 , 6)
b. P(x ≤24)
= P[(x - μ ) / σ (24 - 20) / 6]
= P(z ≤0.67)
= 0.74857
=0.7486
Hence:
Probability = 0.7486
c. P(21 < x < 26)
= P[(21 - 26)/ 6) < (x - μ ) / σ < (24 - 20) / 6) ]
= P(-0.83 < z < 0.67)
= P(z < 0.67) - P(z < -0.)
= 0.74857- 0.2033
= 0.54527
Hence:
Probability =0.54527
d. Using standard normal table ,
P(Z < z) = 66%
P(Z < 0.50) = 0.66
z = 0.50
Using z-score formula,
x = z× σ + μ
x = 0.50 × 6 + 20 = 23
23 Christmas cards
Therefore the distribution of X is X ~ N (20 , 6) and the probability that this American will receive no more than 24 Christmas cards this year is 0.7486.
Learn more about probability here:brainly.com/question/24756209
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