Consider posting each question separately, please.

Given <span>-2.8k-4.31=-15.79 :

1) Isolate the "k" term on the left by adding 4.31 to both sides of this equation.
2) Simplify the right side.
3) Divide both sides by -2.8 and you'll have the answer: k = ?</span>

5 0

3 years ago

You manage a call center that processes orders. One measure of efficiency in your department is the time spent on each call. You

siniylev [52]

Answer:

The answer is option A.

Step-by-step explanation:

If the report is made once a month, the daily variations (including the tendency to decrease efficiency on Friday) will be masked within the monthly result.

It would only generate an ethical problem in the case that Fridays fall in different samples, but in the case of the monthly report there are usually 4 or 5 Fridays included in each sample.

4 0

3 years ago

Below are the results of tossing a number cube 7times. Find the experimental probability of tossing an even number. 6 4 3 2 5 3

Daniel [21]

3/7 or 3:7 because you tossed the dice 7 times and got an even number 3 of those times.

7 0

3 years ago

Read 2 more answers

EXPLAIN why we placed the value of x= 4/3( the minimum value) into the equ of gradient(dy/dx) [in the answer, marking scheme att

aliina [53]

$y=x(x-2)^2$
$\implies y'=(x-2)^2+2x(x-2)=3x^2-8x+4=(3x-2)(x-2)=0$
$\implies x=\dfrac23,x=2$

are the critical points, and judging by the picture alone, you must have $b=\dfrac23$ and $a=2$ . (You might want to verify with the derivative test in case that's expected.)

Then the shaded region has area

$\displaystyle\int_0^2x(x-2)^2\,\mathrm dx=\dfrac43$

I'll leave the details to you.

Now, for part (iv), you're asked to find the minimum of $\dfrac{\mathrm dy}{\mathrm dx}=y'$ , which entails first finding the second derivative:

$y'=3x^2-8x+4$
$\implies y''=6x-8$

setting equal to 0 and finding the critical point:

$6x-8=0\implies x=\dfrac86=\dfrac43$

This is to say the minimum value of $\dfrac{\mathrm dy}{\mathrm dx}$ *occurs when $x=\dfrac43$ *, but this is not necessarily the same as saying that $\dfrac43$ is the actual minimum value.

The minimum value of $\dfrac{\mathrm dy}{\mathrm dx}$ is obtained by evaluating the derivative at this critical point:

$m=\dfrac{\mathrm dy}{\mathrm dx}\bigg|_{x=4/3}=3\left(\dfrac43\right)^2-8\left(\dfrac43\right)+4=-\dfrac43$

4 0

4 years ago