Ask question

Ask question

All categories

3 years ago

5

Calculate the average rate of change for given function, from x=2 to x=7

1 answer:

LenaWriter [7]3 years ago

8 0

"When you calculate the average rate of change of a function, you are finding the slope of the secant line between the two points. f(x) = x2 and f(x + h) = (x + h) Therefore, the slope of the secant line between any two points on this function is 2x + h."

You might be interested in

3w + 7=8 need help with this

LiRa [457]

Hope this helps you out 3w+7=8

The answer is w= 1/3

Step by step:

3w+7=8
Step 1: is subtract is 7 from both sides.
3w+7-7=8-7
3w=1

Step 2: Divide both sides by 3.
3w/3 = 1/3

Your solution is:

w=1/3

6 0

3 years ago

A chemical flows into a storage tank at a rate of (180+3t) liters per minute, where t is the time in minutes and 0<=t<=60

Yuliya22 [10]

Answer:

The amount of the chemical flows into the tank during the firs 20 minutes is 4200 liters.

Step-by-step explanation:

Consider the provided information.

A chemical flows into a storage tank at a rate of (180+3t) liters per minute,

Let $c(t)$ is the amount of chemical in the take at <em>t </em>time.

Now find the rate of change of chemical flow during the first 20 minutes.

$\int\limits^{20}_{0} {c'(t)} \, dt =\int\limits^{20}_0 {(180+3t)} \, dt$

$\int\limits^{20}_{0} {c'(t)} \, dt =\left[180t+\dfrac{3}{2}t^2\right]^{20}_0$

$\int\limits^{20}_{0} {c'(t)} \, dt =3600+600$

$\int\limits^{20}_{0} {c'(t)} \, dt =4200$

So, the amount of the chemical flows into the tank during the firs 20 minutes is 4200 liters.

5 0

3 years ago

Mario sells insurance. His goal is to sell 3 types of insurance to each family on a block. He also wants to sell more than 27 po

qaws [65]

When we <em>divide </em>both sides of the <em>inequality</em>,

3x > 27 by 3, <em>3x(3) > 27(3)</em>

we get x > 9, as the <em>simplified form </em>of the<em> inequality.</em>

Thus, giving us the answer of:

<em>x > 9</em>

8 0

3 years ago

The claim is that the proportion of adults who smoked a cigarette in the past week is less than 0.30, and the sample statistics

IrinaK [193]

Answer: 0.84

Step-by-step explanation:

let p be the population proportion of adults who smoked a cigarette in the past week.

As per given , we have

$H_a: p$

Sample size : n= 1491

The sample proportion of adults smoked a cigarette= $\hat{p}=\dfrac{x}{n}=\dfrac{462}{1491}=0.30985915493\approx0.31$

The test statistic for proportion is given by :-

$z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}$

Substitute all the values , we get

$z=\dfrac{0.31-0.30}{\sqrt{\dfrac{0.30(1-0.30)}{1491}}}\\\\=\dfrac{0.01}{\sqrt{0.000141}}\\\\=\dfrac{0.01}{0.011874342087}=0.84261497731\approx0.84$

Hence, the value of the test statistic = 0.84

6 0

3 years ago

Hello!! I have a question, can you please answer it. ^^

xeze [42]

Answer:

1. 75

2. 1

Step-by-step explanation:

5 0

3 years ago