Answer:
A)what is the probability that the sample mean will be More than 58 pounds
P(x>58)
Formula : 
Substitute the values :
refer the z table
P(x<58)=0.5359
P(X>58)=1-P(x<58)=1-0.5359=0.4641
Hence the probability that the sample mean will be More than 58 pounds is 0.4641
B)what is the probability that the sample mean will be More than 57 pounds
P(x>57)
Formula :
Substitute the values :
refer the z table
P(x<57)=0.5040
P(X>57)=1-P(x<57)=1-0.5040=0.496
Hence the probability that the sample mean will be More than 57 pounds is 0.496
C)what is the probability that the sample mean will be Between 55 and 57 pound
Formula :
Substitute the values :
refer the z table
P(x<57)=0.5040
Formula :
Substitute the values :
refer the z table
P(x<55)=0.4443
P(55<x<57)=P9x<57)-P(x<55) =0.5040-0.4443=0.0597
Hence the probability that the sample mean will be Between 55 and 57 pounds is 0.0597
D)what is the probability that the sample mean will be Less than 53 pounds
Formula :
Substitute the values :
refer the z table
P(x<53)=0.3783
The probability that the sample mean will be Less than 53 pounds is 0.3783
E)what is the probability that the sample mean will be Less than 48 pounds
Formula :
Substitute the values :
refer the z table
P(x<48)=0.2358
The probability that the sample mean will be Less than 48 pounds is 0.2358
Answer:
Answer: A real number that is not rational is called irrational. Irrational numbers include √2, π, e, and φ. The decimal expansion of an irrational number continues without repeating.
Solution would be where the two lines intersect but they never do. They are parallel and never meet
Answer: no solution
The lines never intersect or touch should be the answer.
Hope this helps/works.
The minimum distance is the perpendicular distance. So establish the distance from the origin to the line using the distance formula.
The distance here is: <span><span>d2</span>=(x−0<span>)^2</span>+(y−0<span>)^2
</span> =<span>x^2</span>+<span>y^2
</span></span>
To minimize this function d^2 subject to the constraint, <span>2x+y−10=0
</span>If we substitute, the y-values the distance function can take will be related to the x-values by the line:<span>y=10−2x
</span>You can substitute this in for y in the distance function and take the derivative:
<span>d=sqrt [<span><span><span>x2</span>+(10−2x<span>)^2]
</span></span></span></span>
d′=1/2 (5x2−40x+100)^(−1/2) (10x−40)<span>
</span>Setting the derivative to zero to find optimal x,
<span><span>d′</span>=0→10x−40=0→x=4
</span>
This will be the x-value on the line such that the distance between the origin and line will be EITHER a maximum or minimum (technically, it should be checked afterward).
For x = 4, the corresponding y-value is found from the equation of the line (since we need the corresponding y-value on the line for this x-value).
Then y = 10 - 2(4) = 2.
So the point, P, is (4,2).
