The equation of the line would be y = (1/2)x.
To find the equation of a line that is reflected through the line y = x, we just switch the x and y. Then, solve for y.
y = 2x
x = 2y
0.5x = y
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 = 25235
For the alternative hypothesis,
µ > 25235
This is a right tailed test.
Since the population standard deviation is not given, the distribution is a student's t.
Since n = 100,
Degrees of freedom, df = n - 1 = 100 - 1 = 99
t = (x - µ)/(s/√n)
Where
x = sample mean = 27524
µ = population mean = 25235
s = samples standard deviation = 6000
t = (27524 - 25235)/(6000/√100) = 3.815
We would determine the p value using the t test calculator. It becomes
p = 0.000119
Since alpha, 0.05 > than the p value, 0.000119, then we would reject the null hypothesis. There is sufficient evidence to support the claim that student-loan debt is higher than $25,235 in her area.
Answer:
D
Step-by-step explanation:
First, plot the points they gave you onto a graph (shown in the attached image). The slope is the rise (units up or down from a given point) over the run (units left or right from a given point).
So the slope is 6/4 because to get from the red point to the blue point, you rise 6, and run 4. And from blue to red you rise (remember rise goes up or down) 6 and run 4. 6/4 can be simplified to 3/2 so D is your answer :)
Answer:
a) 10,500 big cats
b) 13,401 big cats
c) 33 years
Step-by-step explanation:
Hi, to answer this question we have to apply an exponential growth function:
A = P (1 + r) t
Where:
p = original population
r = growing rate (decimal form; 5/100= 0.05)
t= years
A = population after t years
Replacing with the values given:
a) At 2005, 1 year passed since 2004 (t=1)
A = 10,000 (1+ 0.05)^1
A = 10,500 big cats
b)
At 2010, 6 years passed since 2004 (2010-2004= 6)
A = 10,000 (1+ 0.05)^6
A = 13,401 big cats
c)
50,000 < 10,000 (1+ 0.05)^t
Solving for t:
50,000/10,000 < 1.05^t
5 < 1.05^t
log 5 < log 1.05^t
log 5 < t ( log 1.05)
log 5 / log 1.05 < t
32.9 years < t
33 years = t
Feel free to ask for more if needed or if you did not understand something.
