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3 years ago

Which expression can be used to determine the average rate of change in f(x) over the interval mc001-1.jpg

2 answers:

8 0

The average rate of change of a function f(x) over an interval (a, b) is given by
$\frac{f(b)-f(a)}{b-a}$

Therefore, given a function, f(x), over an interval (2, 9), the average rate of change of the function can be found using the expression
$\frac{f(9)-f(2)}{9-2}$

IrinaK [193]3 years ago

8 0

Answer:

The answer is D.

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A tennis team has won 15 out of 20 matches they played this season. How many more matches will they need to win in a row to rais

artcher [175]

Answer:

they will need to win 18/20 matches

Step-by-step explanation:

if winning 9/10 matches gives them a winning percentage of 90% then you take 9/10 and multiple it by 2 since they played two times the matches of 10, then you get 18/20

7 0

2 years ago

Need help with this....

Fed [463]

Answer:

m ∠S = 43°

Step-by-step explanation:

Exterior angle = Sum of opposite interior angles

(6x)° = 77° + (2x + 3)°

6x = 77 + 2x + 3

6x - 2x = 77 + 3

4x = 80

x = 20

m ∠S = (2x + 3)° = 2(20) + 3 = 40 + 3 = 43°

4 0

2 years ago

Read 2 more answers

In the past, 35% of the students at ABC University were in the Business College, 35% of the students were in the Liberal Arts Co

lesantik [10]

Answer:

The test statistic for Business College is -0.9944

The test statistic for Liberal Arts College is 1.1483

The test statistic for Education College is 0

Step-by-step explanation:

The percentage of ABC University college students in

Business College = 35%

Liberal Arts College = 35%

Education College = 30%

The number of students in the sample = 300 students

In the sample, the number of students in;

Business College = 90

Liberal Arts College = 120

Education College = 90

Therefore;

The percentage of Business College students = 90/300 × 100 = 30%

The percentage of Liberal Arts College, = 120/300 × 100 = 40%

The percentage of Education College, $\hat p$ = 90/300 × 100 = 30%

The test statistic for the difference in two proportions is given as follows;

$z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p \cdot q}{n}}}$

Where;

$\hat p$ = The sample proportion

p = The population proportion

q = 1 - p

n = The sample size

The test statistic for Business College is;

$z=\dfrac{0.3-0.35}{\sqrt{\dfrac{0.35 \times 0.65}{90}}} \approx -0.9944$

The test statistic for Liberal Arts College is;

$z=\dfrac{0.4-0.35}{\sqrt{\dfrac{0.35 \times 0.65}{120}}} \approx 1.1483$

The test statistic for Education College is;

$z=\dfrac{0.3-0.3}{\sqrt{\dfrac{0.3 \times 0.70}{120}}} \approx 0$

7 0

2 years ago

Lim x->-5(((1)/(5)+(1)/(x))/(10+2x))=<br><br>correct answer 1/10x = -1/50<br><br>explain:

slava [35]

Given:

The limit problem is:

$lim_{x\to -5}\dfrac{\dfrac{1}{5}+\dfrac{1}{x}}{10+2x}$

To find:

The value of the given limit problem.

Solution:

We have,

$lim_{x\to -5}\dfrac{\dfrac{1}{5}+\dfrac{1}{x}}{10+2x}$

It can be written as:

$=lim_{x\to -5}\dfrac{\dfrac{x+5}{5x}}{2(5+x)}$

$=lim_{x\to -5}\dfrac{x+5}{5x}\times \dfrac{1}{2(5+x)}$

$=lim_{x\to -5}\dfrac{1}{5x\times 2}$

$=lim_{x\to -5}\dfrac{1}{10x}$

Applying limit, we get

$lim_{x\to -5}\dfrac{1}{10x}=\dfrac{1}{10(-5)}$

$lim_{x\to -5}\dfrac{1}{10x}=\dfrac{1}{-50}$

Therefore, the value of given limit problem is $-\dfrac{1}{50}$ .

6 0

2 years ago

When preparing an environment for school-aged children, the caregiver should ensure that the

lisov135 [29]

There are no sharp objects around,same with bleach

5 0

3 years ago

Read 2 more answers