Answer: We'll have to integrate the growth rate with respect to time, without forgetting the amount of area covered at the start. A = 2010 + â« 8t1/2 dt (as t goes from 0 to 24) A = 2010 + [(16/3) t3/2] (as t goes from 0 to 24) A = 2010 + 627 = 2637 acres We don't integrate the 2010 here because whatever effects of the passing of time have already taken place over that area. We just treat it as a constant and add it to the result of integration.

8 0

3 years ago

Help me See the picture I’m giving brainliest

Eduardwww [97]

Y=-1.5x + 8

hope this helps :) (please mark brainliest!)

4 0

3 years ago

Andre bought a new bag of cat food. The next day, he opened it to his cat. The graph shows how many ounces was left in the bag o

Viktor [21]

Answer:

There were 16 ounces of Cat Food after 24 days.

Step-by-step explanation:

From the graph attached,

For an ordered pair (x, y) lying on the graph,

x-coordinate represents the number of days since bag was bought and y- coordinate represents the ounces of cat food required.

Now the value x = 24 on x-axis shows the number of days since the bag was bought.

For x = 24, ounces of cat food in the bag ( y ) = 16 ounce

Therefore, 16 ounces of Cat Food were in the bag after 24 days since it was bought.

7 0

3 years ago

Suppose heights of seasonal pine saplings are normally distributed and have a known population standard deviation of 17 millimet

Mumz [18]

Answer:

$296.693\leq x\leq 319.307$

Step-by-step explanation:

The confidence interval for the population mean x can be calculated as:

$x'-z_{\alpha /2}\frac{s}{\sqrt{n} } \leq x\leq x'+z_{\alpha /2}\frac{s}{\sqrt{n} }$

Where x' is the sample mean, s is the population standard deviation, n is the sample size and $z_{\alpha /2}$ is the z-score that let a proportion of $\alpha /2$ on the right tail.

$\alpha$ is calculated as: 100%-99%=1%

So, $z_{\alpha/2}=z_{0.005}=2.576$

Finally, replacing the values of x' by 308, s by 17, n by 15 and $z_{\alpha /2}$ by 2.576, we get that the confidence interval is:

$308-2.576\frac{17}{\sqrt{15} } \leq x\leq 308+2.576\frac{17}{\sqrt{15} }\\308-11.307 \leq x\leq 308+11.307\\296.693\leq x\leq 319.307$

6 0

3 years ago