what is an equation of the line that passes through the point (8,-3) and is perpendicular to the line y-15=6(x+2)

Find the average rate of change from x = 3 to x = 15 for the function f(x) = 0.01(2)x

julsineya [31]

The average rate of change for this is the slope of the secant line that connects those 2 points (3, y) and (15, y). What we need for the slope formula of change in y over change in x are the y values which are unknown as of right now. We can find them though! Don't worry! The equation is y = .01(2)^x. Using that equation, let's sub in both the 3 and the 15 and find the corresponding y values. Subbing in first a 3 gives you y = .01(2)^3, and y = .08. Subbing in a 15 gives you y = .01(2)^15 and y = 327.68. Now we have the coordinates we need to find the slope of the secant line connecting those 2 points: (3, .08) and (15, 327.6). Fitting those into the slope formula gives us (327.68-.08)/(15-3). Simplifying that is 327.6/12 which divides out to 27.3

3 0

3 years ago

1-116 1.2.6

ankoles [38]

Answer:

A. 30 adults; B. 24 not dolls

Step-by-step explanation:

A.

$\begin{array}{rcl}\text{No. of adults} & = & \text{No. of visitors} \times \text{Fraction who are adults}\\& = & \text{100 visitors} \times \dfrac{\text{3 adults}}{\text{10 visitors}} \\& = & \textbf{30 adults}\\\end{array}$

B.

If five-eighths of the prizes were dolls, then three-eighths of the prizes were not dolls.

$\begin{array}{rcl}\text{No. not dolls} & = & \text{No. of prizes} \times \text{Fraction not dolls}\\& = & \text{64 prizes} \times \dfrac{\text{3 not dolls}}{\text{8 prizes}} \\& = & \textbf{24 not dolls}\\\end{array}$

3 0

3 years ago

For a certain river, suppose the drought length Y is the number of consecutive time intervals in which the water supply remains

AnnZ [28]

Answer:

a) There is a 9% probability that a drought lasts exactly 3 intervals.

There is an 85.5% probability that a drought lasts at most 3 intervals.

b)There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation

Step-by-step explanation:

The geometric distribution is the number of failures expected before you get a success in a series of Bernoulli trials.

It has the following probability density formula:

$f(x) = (1-p)^{x}p$

In which p is the probability of a success.

The mean of the geometric distribution is given by the following formula:

$\mu = \frac{1-p}{p}$

The standard deviation of the geometric distribution is given by the following formula:

$\sigma = \sqrt{\frac{1-p}{p^{2}}$

In this problem, we have that:

$p = 0.383$

So

$\mu = \frac{1-p}{p} = \frac{1-0.383}{0.383} = 1.61$

$\sigma = \sqrt{\frac{1-p}{p^{2}}} = \sqrt{\frac{1-0.383}{(0.383)^{2}}} = 2.05$

(a) What is the probability that a drought lasts exactly 3 intervals?

This is $f(3)$

$f(x) = (1-p)^{x}p$

$f(3) = (1-0.383)^{3}*(0.383)$

$f(3) = 0.09$

There is a 9% probability that a drought lasts exactly 3 intervals.

At most 3 intervals?

This is $P = f(0) + f(1) + f(2) + f(3)$

$f(x) = (1-p)^{x}p$

$f(0) = (1-0.383)^{0}*(0.383) = 0.383$

$f(1) = (1-0.383)^{1}*(0.383) = 0.236$

$f(2) = (1-0.383)^{2}*(0.383) = 0.146$

Previously in this exercise, we found that $f(3) = 0.09$

So

$P = f(0) + f(1) + f(2) + f(3) = 0.383 + 0.236 + 0.146 + 0.09 = 0.855$

There is an 85.5% probability that a drought lasts at most 3 intervals.

(b) What is the probability that the length of a drought exceeds its mean value by at least one standard deviation?

This is $P(X \geq \mu+\sigma) = P(X \geq 1.61 + 2.05) = P(X \geq 3.66) = P(X \geq 4)$ .

We are working with discrete data, so 3.66 is rounded up to 4.

Either a drought lasts at least four months, or it lasts at most thee. In a), we found that the probability that it lasts at most 3 months is 0.855. The sum of these probabilities is decimal 1. So:

$P(X \leq 3) + P(X \geq 4) = 1$