First, find the area of the plot:
25*10=250
1000/250 is 4
the population density is 4 squirrels per square foot.
Dat is 4th degree because x is raised to the 4th power and that is the highest power of the placholder
4th degree
Answer:
The distribution is skewed, so use the five-number summary. range: 38, median: 16, half of the data are between 9.5 and 25
Step-by-step explanation:
In the picture attached the histogram is shown. We can see that data is skewed to the right, so we have to use the five-number summary. The range of the data is 39 - 1 = 38 (subtraction of the maximum value to the minimum value); the median is (15 + 17)/2 = 16 (if you order the values, 15 and 17 are in the middle); quartile 1 is 9.25 and quartile 3 is 25.5 (see diagram of box and whisker attached), then half of the data are between those values.
Answer:
The y-intercept of the line would be (C) 8
Step-by-step explanation:
The slope of the line with the equation y = -3x - 5 is -3. When figuring out a slope perpendicular to another line you would use the negative reciprocal of the slope which would lead to the slope of the line to be 1/3.
We know it passes through the point (-3,7), substituting x and y, we would then get our new equation,
.
Solve the equation by multiplying -3 with 1/3 and the product would be -1. Add negative one on both sides and you would get 8 = b.
Answer: a - 4.512 hours
b - 1.94 hours
Step-by-step explanation:
Given,
a) A(t) = 10 (0.7)^t
To determine when 2mg is left in the body
We would have,
A(t) = 2, therefore
2 = 10(0.7)^t
0.7^t =2÷10
0.7^t = 0.2
Take the log of both sides,
Log (0.7)^t = log 0.2
t log 0.7 = log 0.2
t = log 0.2/ 0.7
t = 4.512 hours
Thus it will take 4.512 hours for 2mg to be left in the body.
b) Half life
Let A(t) = 1/2 A(0)
Thus,
1/2 A(0) = A(0)0.7^t
Divide both sides by A(0)
1/2 = 0.7^t
0.7^t = 0.5
Take log of both sides
Log 0.7^t = log 0.5
t log 0.7 = log 0.5
t = log 0.5/log 0.7
t = 1.94 hours
Therefore, the half life of the drug is 1.94 hours