Answer:
a)
b)The p value for this case can be founded like this:
Step-by-step explanation:
We have the following data given:
Number of items (x): 40 30 70 90 50 60 70 40 80 70
Labor (hours) (y): 82 60 139 180 99 119 140 73 157 146
And in order to calculate the correlation coefficient we can use this formula:
For our case we have this:
n=10
So then the correlation coefficient would be r =0.997
Part a
In order to test the hypothesis if the correlation coefficient it's significant we have the following hypothesis:
Null hypothesis:
Alternative hypothesis:
The statistic is gven by:
And is distributed with n-2 degrees of freedom. df=n-2=10-2=8
Part b
The p value for this case can be founded like this:
Is a very low value so we have enough evidence to reject the null hypothesis.
4^y=x
log(4^y)=logx
y*log4=logx
y=logx/log4
so, this is a log function, so it is not B,
vertical asymptote for this function is 0, so it did not move right or left (the graph will be look like log x), correct answer is C
Answer:
22.5
Step-by-step explanation:
First, find the constant of the variation by using the given values of x and y. After that, plug the values of the constant and the new value of x into the equation to find the new value of y.
96-49=47 feet
if the diver swam 96 feet deep...then goes up 49 feet....that means his distance below the lake reduced hence you subtract
Let <em>A(t)</em> denote the amount of salt (in kg) in the tank at time <em>t</em>.
At the start, there are 80 kg of salt in the tank, so <em>A(0)</em> = 80.
Solution flows into the tank at a rate of 8 L/min at a concentration of 0.04 kg/L, so that salt flows in at a rate of
(8 L/min) * (0.04 kg/L) = 0.32 kg/min
Solution flows out at the same rate, but its concentration depends on the amount of salt in the tank. The concentration of the solution is the proportion of salt in the liquid to the total volume of the liquid. Solution flows in and out at 8 L/min, so the volume of liquid (1000 L) stays the same. <em>A(t)</em> is the amount of salt in the tank, so the concentration is <em>A(t)</em>/1000 kg/L. Hence salt flows out at a rate of
(8 L/min) * (<em>A(t)</em>/1000 kg/L) = 0.008 <em>A(t)</em> kg/min
The net rate at which salt flows through the system is then given by the differential equation,
d<em>A(t)</em>/d<em>t</em> = 0.32 - 0.008 <em>A(t)</em>
(Don't forget to include the initial condition)