I think it would be the second graph
For example, in the number 14,509, fourteen thousand five hundred and nine, 1 is in the ten thousands place, and zero is in the tens place.
1 4, 5 0 9
ten thousand^ ten^
Answer:
(a) 0.20
(b) 31%
(c) 2.52 seconds
Step-by-step explanation:
The random variable <em>Y</em> models the amount of time the subject has to wait for the light to flash.
The density curve represents that of an Uniform distribution with parameters <em>a</em> = 1 and <em>b</em> = 5.
So, 
(a)
The area under the density curve is always 1.
The length is 5 units.
Compute the height as follows:


Thus, the height of the density curve is 0.20.
(b)
Compute the value of P (Y > 3.75) as follows:
![P(Y>3.75)=\int\limits^{5}_{3.75} {\frac{1}{b-a}} \, dy \\\\=\int\limits^{5}_{3.75} {\frac{1}{5-1}} \, dy\\\\=\frac{1}{4}\times [y]^{5}_{3.75}\\\\=\frac{5-3.75}{4}\\\\=0.3125\\\\\approx 0.31](https://tex.z-dn.net/?f=P%28Y%3E3.75%29%3D%5Cint%5Climits%5E%7B5%7D_%7B3.75%7D%20%7B%5Cfrac%7B1%7D%7Bb-a%7D%7D%20%5C%2C%20dy%20%5C%5C%5C%5C%3D%5Cint%5Climits%5E%7B5%7D_%7B3.75%7D%20%7B%5Cfrac%7B1%7D%7B5-1%7D%7D%20%5C%2C%20dy%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B4%7D%5Ctimes%20%5By%5D%5E%7B5%7D_%7B3.75%7D%5C%5C%5C%5C%3D%5Cfrac%7B5-3.75%7D%7B4%7D%5C%5C%5C%5C%3D0.3125%5C%5C%5C%5C%5Capprox%200.31)
Thus, the light will flash more than 3.75 seconds after the subject clicks "Start" 31% of the times.
(c)
Compute the 38th percentile as follows:

Thus, the 38th percentile is 2.52 seconds.
Answer:
Step-by-step explanation:
Parallel to y=2÷3x+6
Passes through (6,7)
slope is (2/3)
Solve for when y is 7 and x is 6.
take out the + 6 at the end because it doesnt change the slope. replace with b. (y=mx+b)
replace y and x.
7 = (2/3)(6)+b
solve for b and then put it back into the equatio and make 6 and 7 into x and y again.
Answer:
Option b
Step-by-step explanation:
We have a compound interest problem. With an annual interest rate of 0.675 and an initial payment of 8500, with t = 25 years
Then you must use the annual compound interest formula, which is represented by a growing exponential function:

Where:
h is the interest rate of 0.675
y is the money in the savings account as a function of time
Then substitute the values in the formula and we have:

