Answer:
The lower bound is, 
and the upper bound is 
.
Step-by-step explanation:
Let the random variable <em>X</em> follows a normal distribution with mean <em>μ </em>and standard deviation <em>σ</em>.
The the random variable <em>Z, </em>defined as 
is standardized random variable also known as a standard normal random variable. The random variable 
.
The standard normal random variable has a symmetric distribution.
It is provided that 
.
Determine the upper and lower bound as follows:
![P(-z\leq Z\leq z)=0.51\\P(Z\leq z)-P(Z\leq -z)=0.51\\P(Z\leq z)-[1-P(Z\leq z)]=0.51\\2P(Z\leq z)-1=0.51\\2P(Z\leq z)=1.51\\P(Z\leq z)=0.755](https://tex.z-dn.net/?f=P%28-z%5Cleq%20Z%5Cleq%20z%29%3D0.51%5C%5CP%28Z%5Cleq%20z%29-P%28Z%5Cleq%20-z%29%3D0.51%5C%5CP%28Z%5Cleq%20z%29-%5B1-P%28Z%5Cleq%20z%29%5D%3D0.51%5C%5C2P%28Z%5Cleq%20z%29-1%3D0.51%5C%5C2P%28Z%5Cleq%20z%29%3D1.51%5C%5CP%28Z%5Cleq%20z%29%3D0.755)
Use a standard normal table to determine the value of <em>z.</em>
The value of <em>z</em> such that P (Z ≤ z) = 0.755 is 0.69.
The lower bound is, and the upper bound is .
Answer:
I’m pretty sure it’s 7.5, but I’m not sure! I’m sorry if it’s wrong!
Simplify:
8/36=4/18=2/9=0.22222(repeating)
Your answer would be 2/9 or 0.22222(etc.).
I hope this helps :)
Amount of sales of newspapers for the month of January = $8341.50
Percentage of profit for which the newspaper is sold = 0.5%
Then
Amount of profit made in the month of January = 0.5% * 8341.50 dollars
= (0.5/100) * 8341.50 dollars
= 4170.75/100 dollars
= 41.707 dollars
= 41.71 dollars
So the shop makes a profit of $41.71 in the month of January by selling newspapers worth $8341.50. I hope the procedure is perfectly clear for you to understand.
Answer:
D) 1pi in^3
Step-by-step explanation:
Volume formula for cylinder is:
V=pi*r^2*h
V=pi*1*1
V=pi in^3
