A scientist is growing bacteria in a petri dish. The scatter plot below shows the number of bacteria in the dish over time.
A graph is shown.A graph of 'Time' in days from 0 to 12 on the x axis and 'Number of Bacteria' from 0 to 400 on the y axis is shown. Nine points are plotted beginning at 1 comma 50 increasing to approximately 9 comma 325. Points between these two include approximately 2 comma 110, 3 comma 80, 4 comma 120, 5 comma 175, 6 comma 225, 7 comma 240, and 8 comma 275.
If this growth trend continues, approximately how many bacteria would be in the petri dish on day 10?
this is math my bad i will give free money to who can answer
ever think of typing points as, say, (4,120)?
These points are not too far off a straight line.
There are several handy linear regression calculators online you can use to get the line of best fit.
Then just plug in x=10 for that equation.
But, since the differences are approximately 50 for each point, and the overall slope of the line is 325/9 = 36.1
Maybe just guess y(10) = 370 or so
To estimate the number of bacteria in the petri dish on day 10, we can use the scatter plot provided.
Let's observe the trend in the scatter plot. As time (x-axis) increases, the number of bacteria (y-axis) also increases.
By visually inspecting the scatter plot, we can see that the relationship between time and the number of bacteria appears to be approximately linear. The points on the scatter plot form a roughly straight line that is sloping upwards.
To estimate the number of bacteria on day 10, we can draw a line of best fit through the scatter plot points and use it to make our estimate.
Based on the points given, the line of best fit would pass through approximately (9, 325) and (10, ?). We can estimate the y-coordinate (number of bacteria) for day 10 by extending the line of best fit and finding the corresponding point on the y-axis.
From the scatter plot, we can see that the line seems to have a slope of around (325 - 50)/(9 - 1) = 275/8 = 34.375.
Using this estimated slope, we can approximate the number of bacteria on day 10 as follows:
Gradient of the line: 34.375
Let's assume the initial number of bacteria is 50 (based on the first point).
Using the slope-intercept form of a linear equation (y = mx + b), we can find the y-intercept (b) by substituting one of the points we have (e.g., (1, 50)) into the equation.
50 = 34.375 * 1 + b
50 = 34.375 + b
b ≈ 50 - 34.375
b ≈ 15.625
So, the equation of the line is approximately y = 34.375x + 15.625.
Now, let's substitute x = 10 into the equation to find the approximate number of bacteria on day 10:
y ≈ 34.375 * 10 + 15.625
y ≈ 343.75 + 15.625
y ≈ 359.375
Therefore, based on the estimated trend, there would be approximately 359 bacteria in the petri dish on day 10.