Answer:
t = time in decimal years; e.g., 6 months is calculated as 0.5 years. Divide your partial year number of months by 12 to get the decimal years.
Step-by-step explanation:
Answer:
1. first one
2. there are no solutions that i see
Step-by-step explanation:
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.
That means when x=0
e^-0.2x =e^0=1
So f(0)=15/5=3
0 and 2
f(x) = g(x) will be the input or x value at which f and g have the same output or y value. Look in the table where two numbers repeat right next to each other.
−1 −7/2 −9/2
0 −3 −3
1 −2 −3/2
2 0 0
3 4 3/2
4 12 3
5 28 9/2
There are two solutions to f(x) = g(x) which are x=0 and x=2.
