If the true mean is 50 and we reject the hypothesis that μ = 50, what is the probability of Type II error? Hint: This is a trick question.
Answers
Answer:
Zero
Step-by-step explanation:
Given that:
the true mean is 50 and we reject the hypothesis that μ = 50
The probability of the Type II error will be zero, given that we reject the null hypothesis. This has nothing to do with if it is true or false.
Type II error is occurs when you accept a false null hypothesis. The probability of this error is denoted by beta which relies on sample size and population variance.
The probability of rejecting is equal to one minu beta. i.e it is the researchers goal to reject a false null hypothesis.
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The probability that an adult possesses a credit card is .70. A researcher selects two adults at random. By assuming the independence, the probability that the first adult possesses a credit card and the second adult does not possess a credit card is:
A mountain bike costs $675. If the sales tax is 8%, find the total cost of the bike. Round to the nearest cent, ifnecessary
Answers
Answer:
y=6x-8
Step-by-step explanation:
10-4/3-2
m=6/1
y=mx+b
y=6x+b
4=6(2)+b
4=12+b
-8=b
y=6x-8
Answers
Answer:
a^7
Step-by-step explanation:
Answers
Answer:
27
Step-by-step explanation:
First is 9 / 3 which is 3 and then 4 x 6 which is 24 then 24 + 3 = 27
Answer:
PEMDAS
4 × 6 is 24
9 ÷ 3 is 3
24 + 3 = 27
Step-by-step explanation:
Answers
Answer:
The value of x is: 2.5 units
Step-by-step explanation:
Two triangles are said to be similar if the ratio of the corresponding sides of the two triangles are equal.
i.e. if two triangles ΔABC and ΔDEF are similar such that the sides of the triangle ABC are a, b and c and the corresponding sides in ΔDEF are d,e and f respectively then we have:
Here we have the base length of the orange i.e. the quadrant above the x-axis as: 8 units
and the base length of the similar triangle i.e. triangle below x-axis as: 4 units.
i.e. we have: a=8 and d=4
and b=5 and e=x
Hence, we have:
i.e.
Hence, we have:
Answers
Given:
The statement is "the difference of 18 and 12".
To find:
The expression for the given statement.
Solution:
We have,
Given statement = The difference of 18 and 12.
We use subtraction, to find the different between two numbers and subtraction is represented by negative sign. So, we need to subtract second number form the first number.
The difference of 18 and 12 = 18 - 12
Therefore, the required expression for the given statement is 18-12.
a. Formulate an LP model for this problem.
b. Sketch the feasible region.
c. What is the optimal solution?
Answers
Answer:
Let X1 be the number of decorative wood frame doors and X2 be the number of windows.
The profit earned from selling each door is $500 and the profit earned from selling of each window is $400.
The Sanderson Manufacturer wants to maximize their profit. So for this model, the objective function is
Max: 500X1 + 400X2
Now the total time available for cutting of door and window are 2400 minutes.
so the time taken in cutting should be less than or equal to 2400.
60X1 + 30X2 ≤ 2400
The total available time for sanding of door and window are 2400 minutes. Therefore, the time taken in sanding will be less than or equal to 2400. 30X1 + 45X2 ≤ 2400
The total time available for finishing of door and window is 3600 hours. Therefore, the time taken in finishing will be less than or equal to 3600. 30X1 + 60X2 ≤ 3600
As the number of decorative wood frame door and the number of windows cannot be negative.
Therefore, X1, X2 ≥ 0
so the questions
a)
The LP mode for this model is;
Max: 500X1 + 400X2
Subject to:
60X1 + 30X2 ≤ 2400
]30X1 +45X2 ≤ 2400
30X1 + 60X2 ≤ 3600
X1, X2 ≥ 0
b) Plot the graph of the LP
Max: 500X1+ 400X2
Subject to:
60X1 + 30X2 ≤ 2400
30X1 + 45X2 ≤ 2400
30X1 + 60X2 ≤ 3600
X1,X2
≥ 0
In the uploaded image of the graph, the shaded region in the graph is the feasible region.
c) Consider the following corner point's (0,0), (0, 53.33), (20, 40) and (40, 0) of the feasible region from the graph
At point (0, 0), the objective function,
500X1 + 400X2 = 500 × 0 + 400 × 0
= 0
At point (0, 53.33), the value of objective function,
500X1 + 400X2 = 500 × 0 + 400 × 53.33 = 21332
At point (40, 0), the value of objective function,
500X1 + 400X2 = 500 × 40 + 400 × 0 = 20000
At point (20, 40), the value of objective function
500X1 + 400X2 = 500 × 20 + 400 × 40 = 26000
The maximum value of the objective function is
26000 at corner point ( 20, 40 )
Hence, the optimal solution of this problem is
X1 = 20, X2 = 40 and the objective is 26000