Please help me i really need help!!!

After an exposure incident, who gives a written opinion of the incident to the facility?

Ivenika [448]

The doctor gives a written opinion of the incident to the facility.

3 0

3 years ago

A teacher collected information from a class of 25 students about the time, in hours, they spent studying the previous week and

djverab [1.8K]

Answer:

B. −0.72

Explanation:

The correlation coefficient is a measure of the linear association between two variables.

It is a dimensionless value bettween - 1 and +1. Hence, it does not depend on the units that are used to measure the data.

You can see the formula to calculate the correlation coefficient in the attached picture.

There you see that the numerator containts the sum of the differences of the values of the x-variable with its the mean, multiplied by the sum of the differences of the values of the y variable with its mean, while the denominator contains the square root of the product of the squares of the same differences, thus the units are canceled.

In conclusion, If the teacher changes the units of each variable from hours to minutes, the value of the correlation coefficient between minutes studying and minutes spent on the Internet will not change; it will continue being −0.72.

5 0

3 years ago

HEYY I NEED HELP WITH AN AP CALC ASSIGNMENT ASAP

Vladimir79 [104]

CALCULATOR PART

1. The area of R + S is unsigned, meaning you want to find

$\displaystyle\int_a^b\left|f(x)-g(x)\right|\,\mathrm dx$

where $[a,b]$ is the interval between the leftmost and rightmost intersections of $f$ and $g$ .

First use your calculator to find these intersections:

$\cos x=\dfrac{x+1}3\implies x\approx-3.64,x\approx-1.86,x\approx0.889$

so that $a=-3.64$ and $b=0.889$ . Now compute the integral using your calculator:

$\displaystyle\int_a^b\left|f(x)-g(x)|\,\mathrm dx\approx1.662$

2. The volume, using the washer method, is given by the integral

$\displaystyle\pi\int_{-1.86}^{0.889}(|2-g(x)|^2-|2-f(x)|^2)\,\mathrm dx\approx12.078$

3. A circle of radius $r$ has area $\pi r^2$ ; a semicircle with the same radius thus has area $\frac{\pi r^2}2$ . Each cross section of this solid is a semicircle whose diameter is the vertical distance between $f(x)$ and $g(x)$ , or $|f(x)-g(x)|$ . In terms of the diameter $d=2r$ , the area of each semicircle would be $\frac{\pi d^2}8$ . Then the volume of the solid is

$\displaystyle\frac\pi8\int_{-3.64}^{-1.86}|f(x)-g(x)|^2\,\mathrm dx\approx0.0425$

NON-CALCULATOR PART

4. The mean value theorem says that for a function $F$ continuous on an interval $[a,b]$ and differentiable on $(a,b)$ , there is some $c\in(a,b)$ such that

$F'(c)=\dfrac{F(b)-F(a)}{b-a}$

If this $F$ happens to be an antiderivative of $f$ , then we end up with

$f(c)=\displaystyle\frac1{b-a}\int_a^bf(x)\,\mathrm dx$

$\cos x$ is continuous and differentiable everywhere, so the MVT applies. We have $F'(x)=f(x)=\cos x$ , so the MVT tells us there is some $c\in[0,\pi$ such that

$\cos c=\dfrac{\sin\pi-\sin0}{\pi-0}=0$

That is, the average value of $f(x)$ on $[0,\pi]$ is 0. The MVT says there is some $c$ in the interval such that the function takes on the average value itself; this happens for $c=\frac\pi2$ .