If O is an angle in standard position and its terminal side passes through the point

What is the slope of a line perpendicular to the line whose equation is 2x-y=5 ?

mariarad [96]

Answer:

-1/2

Step-by-step explanation:

the slope of the line 2x-y=5 is 2

a perpendicular line has the negative reciprocal of the slope of the other line

hopefully this helps :)

3 0

3 years ago

a boat capsized and sank in the lake. based on an assumption of a mean weight of 143 lb, the boat was rated to carry 70 passenge

Alenkinab [10]

The final answer is the probability that the boat is overloaded because the mean weight of the 70 passengers is greater than 143lb is 1.0000.

A normal distribution is a continuous data distribution with a bell-shaped curve. The normally distributed random variable X has mean $\mu$ and standard deviation $\sigma$ .

In addition, the standard normal distribution represents a normal curve with a mean of zero and a standard deviation of one.

The number of standard deviations an element deviates from the mean is indicated by a standardized Z-score.

What is the theorem of the central limit?

If the sample size from the population is sufficiently large and has a finite variance, the mean of all samples taken will be approximately the same as the population mean. And the variance would be the population variance divided by the sample size. In addition, the sampling distribution would be roughly normal.
The formula for sampling distribution of the sample mean is $\mu_{\bar{x}}=\mu$
The formula for standard deviation of the sample mean is, $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$
Here, $\sigma$ is the population standard deviation and n is the sample size.
The formula for standard z-score is,

$z=\frac{\bar{x}-\mu_{x}}{\frac{\sigma}{\sqrt{n}}}$

From the given information,Population mean, $\mu=171$ ,Population standard deviation, $\sigma=37.7$ ,Sample size, n=70
Compute the probability that the boat is overloaded because the mean weight of the 70 passengers is greater than $143l \mathrm{~b}$ .

$\begin{aligned}P(\bar{X} > 143) &=P\left(\frac{\bar{X}-\mu}{\sigma / \sqrt{n}} > \frac{143-171}{37.7 / \sqrt{70}}\right) \\&=P(Z > -6.21) \\&=1-P(Z < -6.21) \text { (Use the Excel function NORMSDIST(-6.21)) } \\&=1-0.0000 \\&=1.0000\end{aligned}$