The slope of the line is - 3/4

I got my ordered pairs by finding two points on the lines, an noting down their X and Y values.
Then I used the equation Y2 - Y1 (all over) X2 - X1 by plugging in the ordered pairs.
Finally, I solved and got -3/4!
See the attached picture below

Additionally, you can use rise over run, and count the rise and the run. In the image, the rise is three, and the run is four. The slope is negative, which leaves me to know that the answer is -3/4.

Hope this helps :)

7 0

3 years ago

The amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.2 minutes and

Ksju [112]

Answer:

A) 1

B) 1

C) 0

Step-by-step explanation:

The amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.2 minutes and standard deviation 1.5 minutes. Suppose that a random sample of n = 49 customers is observed. Find the probability that the average time waiting in line for these customers is A) Less than 10 minutes B) Between 5 and 10 minutes C) Less than 6 minutes

We solve this question using the z score formula

z = (x-μ)/σ/√n, where

x is the raw score

μ is the population mean

σ is the population standard deviation.

n = number of random samples

A) Less than 10 minutes

x < 10

z = 10 - 8.2/ 1.5 / √49

z = 8.4

P-value from Z-Table:

P(x<10) = 1

B) Between 5 and 10 minutes

For x = 5 minutes

z = 5 - 8.2/ 1.5 / √49

z = -14.93333

P-value from Z-Table:

P(x = 5) = 0

For x = 10 minutes

z = 10 - 8.2/ 1.5 / √49

z = 8.4

P-value from Z-Table:

P(x = 10) = 1

The probability that the average time waiting in line for these customers is between 5 and 10 minutes

P(x = 10) - P(x = 5)

= 1 - 0

= 1

C) Less than 6 minutes

x < 6

z = 6 - 8.2/ 1.5 / √49

z = -10.26667

P-value from Z-Table:

P(x<6) = 0

6 0

3 years ago

Solve problem by dividing 39÷2886= ​

Solve problem by dividing 39÷2886=