Answer:
b
Step-by-step explanation:
Answer:
C. (1, 10) and (6, 5)
Step-by-step explanation:
To find the point with a distance of 5 units.
Let the points be x and y.
x = x1 and x2
y = y1 and y2
x2 - x1 = 5 units
y2 - y1 = 5 units.
A quick look at the options, we can tell that the correct option is C.
From points (1, 10) and (6, 5)
x1 = 1, x2 = 6, y1 = 10 and y2 = 5
Substituting into the equation above, we have;
x2 - x1 = 6 - 1 = 5 units
y2 - y1 = 5 - 10 = -5 units.
Therefore, option C with points (1,10) and (6,5) have a distance of five units between them.
Answer:
a) 
b) 
c) 
Step-by-step explanation:
If we work with the limits defined from 5 to 10 then part b and c from this question not makes sense. If we work with the limits 1 and 6 all the parts for the question makes sense because if we work with 5 and 10 the only thing that we can find is the expected value 
Assuming the following correct question : "A publication released the results of a study of the evolution of a certain mineral in the Earth's crust. Researchers estimate that the trace amount of this mineral x in reservoirs follows a uniform distribution ranging between 1 and 6 parts per million"
Solution to the problem
Let A the random variable that represent " amount of the mineral x ". And we know that the distribution of A is given by:

Part a
For this uniform distribution the expected value is given by
where X is the random variable, and a,b represent the limits for the distribution. If we apply this for our case we got:

Part b
For this case we can use the cumulative distribution function for the uniform distribution given by:

And we want this probability:
Part c
For this case we want this probability:

Answer:
Step-by-step explanation:
Given that for a sample of eight bears, researchers measured the distances around the bears' chests and weighed the bears.
r = linear correlation coeff = 0.952
H_0: r =0\\
H_1 :r\neq 0
(Two tailed test)
r difference = 0.952
n=8
Std error = \sqrt{\frac{1-r^2}{n-2} } =0.12496
Test statistic t = 0.952/0.12496 = 7.618
Alpha = 0.05
df = 6
p value = 0.000267
This implies H0 is rejected.
There exists a linear relation between the variables and r cannot be 0
0.952^2 = 0.906=90.6% of variation in weight can be explained by the linear relationship between weight and chest size
Money= how many hours
$4= 1h
$8=2h
$12=3h
$16=4h
$20=5h
$24=6h
l hope that helps :)