Answers...........................?

The concentration C (in mg/dl), of an antibiotic in a patient’s bloodstream per hour, t, is given by: c(t) = (50t)/(t^2+25) In o

NARA [144]

Answer:

Step-by-step explanation:

<u>Functions </u>

The problem describes a function that expresses the concentration of an antibiotic in mg/dl vs time in hours as:

$\displaystyle c(t)=\frac{50\ t}{t^2+25}$

We need to find the first value of t such that

$\displaystyle c(t)\geq 4$

It means that

$\displaystyle \frac{50\ t}{t^2+25}\geq 4$

Operating with the inequality

$\displaystyle 50\ t\geq 4\ t^2+100$

Rearranging and dividing by 2, we have a polynomial inequality:

$\displaystyle 2t^2-25t+50\leq 0$

Factoring

$\displaystyle 2(t-10)\left (t-\frac{5}{2}\right )\leq 0$

There are two possible values for t, both valids because they are positive

$\displaystyle t=\frac{5}{2}=2.5, \ t=10$

We need to find the first value, i.e.

$t=2.5 \ hours$

Now for the graphic method, we plot the graph for the function and a horizontal line at c=4 to find the values of t.

The graph is shown in the image provided below. We can see both values where the funcion and C=4 intersect. Both values coincide with the previously analitically found values

4 0

3 years ago

If X and Y are any random variables with E(X) = 5, E(Y) = 6, E(XY) = 21, V(X) = 9 and V(Y) = 10, then the relationship between X

natka813 [3]

Answer:

We have a strong negative relationship between the variables.

Step-by-step explanation:

Given two random variables X and Y, it is possible to calculate the covariance as Cov(X, Y) = E(XY)-E(X)E(Y). We have E(X)=5, E(Y)=6 and E(XY)=21. Therefore Cov(X,Y)=21-(5)(6)=21-30=-9. On the other hand, we know that the correlation of X and Y is the number defined by $Cov(X,Y)/\sqrt{Var(X)}\sqrt{Var(Y)}$ and because in this particular case we have V(X)=9 and V(Y)=10, we have $-9/\sqrt{9}\sqrt{10}$ = -0.9487. Therefore, we have a strong negative relationship between the variables.