Answer:
The probability that sample mean differ the true greater than 2.1 will be 2.8070 %
Step-by-step explanation:
Given:
Sample mean =46 dollars
standard deviation=8
n=53
To Find :
Probability that sample mean would differ from true mean by greater than 2.1
Solution;
<em>This sample distribution mean problem,</em>
so for that
calculate Z- value
Z=(sample mean - true mean)/(standard deviation/Sqrt(n))
Z=-2.1/(8/Sqrt(53))
Z=-2.1*Sqrt(53)/8
Z=-1.91102
Now for P(X≥2.1)=P(Z≥-1.91102)
Using Z-table,
For Z=-1.91
P(X>2.1)=0.02807
Answer:
a-bi
Step-by-step explanation:
If a quadratic equation lx^2+mx+n=0 has one imaginary root as a+bi then the other root is the conjugate of a+bi = a-bi
Because we have l, m and n are real numbers and they are the coefficients.
Sum of roots = a+bi + second root = -m/l
When -m/l is real because the ratio of two real numbers, left side also has to be real.
Since bi is one imaginary term already there other root should have -bi in it so that the sum becomes real.
i.e. other root will be of the form c-bi for some real c.
Now product of roots = (a+bi)(c-bi) = n/l
Since right side is real, left side also must be real.
i.e.imaginary part =0
bi(a-c) =0
Or a =c
i.e. other root c-bi = a-bi
Hence proved.
-8
Step 1) Subtract 3 from both sides of the equation
Step 2) Add 4W to both sides of the equation
Step 3) Divide both sides of the equation by the same factor
1. x^15
2. m^40
3. v^7/2
4. k^4
5. 1/x^14
6. 1/r^3/2
7. b^25
8. h^2
9. m^3n^1/7
10. x^12
11. 1/g^37
12. 1/100m^6
13. j^6/216
14. 1/125f^3
15. 3z^1/2
16. 1/100m^6
17. j^6/216
18.1/81d^20
19. 1
20. g^1/2 r^3
21. 16a^11
22. m^16n^21
23. 1
24. 1/y^10x^6
25. 343s^10/t^17/2
26.n^17/2 /m^10/2
27. 729/b^24c^9
28. 20y^2/x^33
7,000 and 7 hope this helped
