Someone help on the first and number

Find the average rate of change of the function between the given values of x. y = 9+ 5x + 0.5x2 between x = 2 and x = 4.

svet-max [94.6K]

Answer:

The average rate of change is 8.

Step-by-step explanation:

The formula to calculate the average rate of change of a function F(x) is:

$\frac{F(b)-F(a)}{b-a}$

In this case, F(x) = $0.5x^{2} +5x+9$

a=2 and b=4

You have to evaluate x=2 (which is a in the formula) and x=4 (which is b in the formula) in the function.

In order to obtain F(b) and F(a) you have to replace x=4 and x=2 in the given function:

F(b) = $(0.5)4^{2} + 5(4) +9= 37$

F(a) = $(0.5)2^{2} + 5(2)+9=21$

$\frac{F(b)-F(a)}{b-a} = \frac{37-21}{4-2}=\frac{16}{2} = 8$

The answer is 8.

3 0

2 years ago

Total profit P is defined as total revenue R -= total cost C, and is given by the function Upper P left parenthesis x right par

Goshia [24]

Answer:

Step-by-step explanation:

P(x) = R(x) - C(x)

Where,

P(x) = Total profit

R(x) = Total revenue

C(x) = Total cost

R(x)equals150.97 x minus 0.3 x squared

R(x) = 150.97x - 0.3x^2

Upper C left parenthesis x right parenthesis equals 4529.10 plus 0.5 x squared.

C(x) = 4529.10 + 0.5x^2

P(x) = R(x) - C(x)

= (150.97x - 0.3x^2) - ( 4529.10 + 0.5x^2)

= 150.97x - 0.3x^2 - 4529.10 - 0.5x^2

= 150.97x - 0.8x^2 - 4529.10

P(x) = -0.8x^2 + 150.97x - 4529.10

4 0

3 years ago

(-8x) divided by (-2)

Ghella [55]

4x is your answer to your question

6 0

3 years ago

Read 2 more answers

Find the equation of the line that is perpendicular to y=-2x-9 and contains the points (8,-4)

avanturin [10]

The slope of the given line is -2, and since perpendicular lines have negative reciprocal slopes, the slope of the line we want to find is 1/2.

Substituting into point-slope form,

$y+4=\frac{1}{2}(x-8)\\\\y+4=\frac{1}{2}x-4\\\\\boxed{y=\frac{1}{2}x-8}$

6 0

1 year ago

As part of the Pew Internet and American Life Project, researchers conducted two surveys in late 2009. The first survey asked a

REY [17]

Answer:

The 95% confidence interval for the difference between the proportion of all U.S. teens and adults who use social networking sites is (0.223, 0.297). This means that we are 95% sure that the true difference of the proportion is in this interval, between 0.223 and 0.297.

Step-by-step explanation:

Before building the confidence interval we need to understand the central limit theorem and the subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Sample of 800 teens. 73% said that they use social networking sites.

This means that:

$p_T = 0.73, s_T = \sqrt{\frac{0.73*0.27}{800}} = 0.0157$

Sample of 2253 adults. 47% said that they use social networking sites.

This means that:

$p_A = 0.47,s_A = \sqrt{\frac{0.47*0.53}{2253}} = 0.0105$

Distribution of the difference:

$p = p_T - p_A = 0.73 - 0.47 = 0.26$

$s = \sqrt{s_T^2+s_A^2} = \sqrt{0.0157^2+0.0105^2} = 0.019$

Confidence interval:

Is given by:

$p \pm zs$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

Lower bound:

$p - 1.96s = 0.26 - 1.96*0.019 = 0.223$

Upper bound:

$p + 1.96s = 0.26 + 1.96*0.019 = 0.297$

The 95% confidence interval for the difference between the proportion of all U.S. teens and adults who use social networking sites is (0.223, 0.297). This means that we are 95% sure that the true difference of the proportion is in this interval, between 0.223 and 0.297.

3 0

2 years ago