Answer:
Step-by-step explanation:
We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean
For the null hypothesis,
µ = 17
For the alternative hypothesis,
µ < 17
This is a left tailed test.
Since the population standard deviation is not given, the distribution is a student's t.
Since n = 80,
Degrees of freedom, df = n - 1 = 80 - 1 = 79
t = (x - µ)/(s/√n)
Where
x = sample mean = 15.6
µ = population mean = 17
s = samples standard deviation = 4.5
t = (15.6 - 17)/(4.5/√80) = - 2.78
We would determine the p value using the t test calculator. It becomes
p = 0.0034
Since alpha, 0.05 > than the p value, 0.0043, then we would reject the null hypothesis.
The data supports the professor’s claim. The average number of hours per week spent studying for students at her college is less than 17 hours per week.
{(2, 3), (1, 5), (2, 7)} - NOT because 2->3 and 2->7
{(11, 9), (11, 5), (9, 3)} - NOT because 11->9 and 11->5
{(3, 8), (0, 8), (3, -2)}
- NOT because 3->8 and 3->-2
{(-1, 5), (-2, 6), (-3, 7)} - YES
Answer:
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Answer:
10
Step-by-step explanation:
Given expression is 
.
is a complex expression.
The value of<em> i </em>is –1.
By definition of absolute value, 
Now, substitute in the above formula.
(since <em>i </em>= –1)
(since 
and 
)
(Square root of 100 is 10)
= 10
Hence, the value of the expression is 10.
Answer: the answer is 0.435
Step-by-step explanation:
