Answer:
The answer is false
Step-by-step explanation:
In a sample above 30 obs like this the confidence interval is defined as
X+- t* (s/sqrt(n)) where X is the mean t the tvalue for a given confidence level, n the size of sample and s standar deviation.
To find de appropiate value of t we must see the T table where rows are degrees of freedom and columns significance level
The significance is obtained:
significance = 1 - confidence level = 1 - 0.9 = 0.10
Degrees of freedom (df) for the inteval are
df = n - 1 = 18 - 1 = 17
So we must look for the value of a t with 17 values and significance of 0.10 which in t table is 1.740 not 1.746 ( thats the t for 16 df)
I would go with Jesses survey.
Unlike everyone else’s survey Jesse chose 6th grade students in random, causing a more unbiased research.
Answer:
Step-by-step explanation:
The middle line is the median, then from left to right the lines are the first, second, third and fourth quartile.
Answer:
Step-by-step explanation:
From the given information:
a) To express the weekly profit as a function of price
Cost =C(q) = 1500 + 10q
Revenue = p×q = (50 − 0.1q)×q = 50q - 0.1q²
Revenue = 50q - 0.1q²
Weekly profit = Revenue - Cost
P(q) = (50q -0.1q²) - (1500 + 10q)
P(q)= -0.1 q² + 40 q - 1500
However, q = 500 - 10 p using p = 50 − 0.1q
P= -0.1 (500 - 10 p)² + 40 (500 - 10 p) - 1500
P= -10 p² + 600 p - 6500
b)
The price at which the bottle of the wine must be sold to realise a maximum profit can be determined by finding the derivative and then set it to 0
P' = 0
= -20p+600 = 0
20p = 600
p = 600/20
p = $30
c)
The maximum profit that can be made by the producer is:
P= -10(30)² + 600(30) - 6500
P = - 9000 + 18000 - 6500
P = $2500