Answer:7+a
Step-by-step explanation:multiply 7 by a
Answer:
(1) Correct option is B.
(2) Correct option is C.
Step-by-step explanation:
The information provided is:

The (1 - <em>α</em>)% confidence interval for the difference between two mean is:

The critical value of <em>t</em> is:

degrees of freedom 

Compute the 95% confidence interval for the difference between two mean as follows:

Thus, the 95% confidence interval, (2.14, 3.86) implies that the true mean difference value is contained in this interval with probability 0.95.
Correct option is B.
The null value of the difference between means is 0.
As the value 0 is not in the interval this implies that there is a difference between the two means, concluding that priming does have an effect on scores.
Correct option is C.
Area ( If you know the circumference) :
C^2 / (4×3.14) Brackets are important!
So you have to type :
53.38^2/(4×3.14)
= 226.865
The answer is 69.6 degrees because all of the angles of a triangle always add up to 180 degrees so adding 42 and 68.4 gets 110.4, and then subtracting that from 180 gets 69.9.
Answer:
0.0903
Step-by-step explanation:
Given that :
The mean = 1450
The standard deviation = 220
sample mean = 1560



P(X> 1560) = P(Z > 0.5)
P(X> 1560) = 1 - P(Z < 0.5)
From the z tables;
P(X> 1560) = 1 - 0.6915
P(X> 1560) = 0.3085
Let consider the given number of weeks = 52
Mean
= np = 52 × 0.3085 = 16.042
The standard deviation =
The standard deviation = 
The standard deviation = 3.3306
Let Y be a random variable that proceeds in a binomial distribution, which denotes the number of weeks in a year that exceeds $1560.
Then;
Pr ( Y > 20) = P( z > 20)


From z tables
P(Y > 20)
0.0903