Given:
n = 48, the sample size
m = $25,000, th sample mean
σ = $3000, the maximum variance
The standard deviation is
s = √σ = √(3000) = 54.772
As a rule of thumb, the range is 4 times the standard deviation. Therefore the range is
R = 4*54.772 = 219.09
Half the range is
R/2 = 219.09/2 = 109.45
The required range is
(25000-109.45, 25000+109.45) = (24890.46, 25109.54)
Answer: ($24,890.46, $25,109.54)
Well this is were you would need to multiply the amount of pounds of rice they did and then divide the amount of months they give you!
Answer:
it minimum
Step-by-step explanation:
you didn't really give an equation
a. Find the probability that an individual distance is
greater than 214.30 cm
We find for the value of z score using the formula:
z = (x – u) / s
z = (214.30 – 205) / 8.3
z = 1.12
Since we are looking for x > 214.30 cm, we use the
right tailed test to find for P at z = 1.12 from the tables:
P = 0.1314
b. Find the probability that the mean for 20 randomly
selected distances is greater than 202.80 cm
We find for the value of z score using the formula:
z = (x – u) / s
z = (202.80 – 205) / 8.3
z = -0.265
Since we are looking for x > 202.80 cm, we use the
right tailed test to find for P at z = -0.265 from the tables:
P = 0.6045
c. Why can the normal distribution be used in part (b),
even though the sample size does not exceed 30?
I believe this is because we are given the population
standard deviation sigma rather than the sample standard deviation. So we can
use the z test.
Answer:
1. Use the distance formula and midpoint formula
2a. Learn the properties of an orthocenter
2b. Write a conjecture and then verify if it's true for acute and right triangles
3. Prove that D is not equidistant from A to C
Step-by-step explanation:
