Answer:
D. They do not have the same ratio of girls to boys because a straight line cannot be drawn through the points (14, 16), (17, 19), and (0, 0) on a coordinate grid.
Step-by-step explanation:
In order to check if the ratio is the same, convert the ratios to (x, y) points, plot them on a graph and see if you can draw a straight line through them and the origin (0, 0). If you can, then the ratios are the same. If not, then the ratios are not the same.
After plotting (14, 16) and (17, 19) and (0, 0) on a graph (attached), you can see that a straight line will only go through any two of the points, but not all three. Therefore, the ratios are not the same.
Answer:
Bus B travels faster.
Step-by-step explanation:
Bus A travels according to the equation
y = 125/2 x
This equation is in the form y = mx + b, where m = 125/2, and b = 0.
m is the slope, so the slope is 125/2 which equals 62.5
The slope is the rate of change. Since y is distance in miles, and x is time in hours, the slope is the speed in miles per hour.
The speed of Bus A is 62.5 miles per hour.
Bus B travels according to the graph. We can find the slope by using easy-to-read points on the graph. The slope is the speed in miles per hour.
Look for two points on grid line intersections since they are the easiest ones to read. Let's use point (0, 0) and point (3, 200). Now that we have 2 points, we can find the slope.
![slope = \dfrac{y_2 - y_1}{x_2 - x_1}](https://tex.z-dn.net/?f=%20slope%20%3D%20%5Cdfrac%7By_2%20-%20y_1%7D%7Bx_2%20-%20x_1%7D%20)
![slope = \dfrac{200 - 0}{3 - 0}](https://tex.z-dn.net/?f=%20slope%20%3D%20%5Cdfrac%7B200%20-%200%7D%7B3%20-%200%7D%20)
![slope = \dfrac{200}{3} = 66.67](https://tex.z-dn.net/?f=%20slope%20%3D%20%5Cdfrac%7B200%7D%7B3%7D%20%3D%2066.67%20)
The speed of Bus 2 is 66.67 miles per hour.
Answer: Bus B travels faster.
35.2ft
You can get this my multiplying the scale factor by the height. In this case you need to use the bottom number to multiply since you are going in reverse.
Answer: 20w + 3
———————
10
Step-by-step explanation:
We know that
<span>This problem can be represented through the following equation
</span>
A = A₀(1/2)^t/h
where
A-----------> is the amount of substance left after time t
A₀ ----------> is the original amount---------> 2 g
h-------------> is the half-life-----------> 8 days
for A=0.04 g
t=?
0.04 = 2(1/2)^t/8
0.02 = (1/2)^t/8
Take ln on both sides:
ln(0.02) = ln [(1/2)^t/8]
ln(0.02) = (t/8)(ln 1/2)
t = 8*ln(0.02)/ln(1/2)
t = 45.15 days
the answer is 45.15 days