1) Find the graph of a line passing through (-1, 4) and (2, 0).
The slope of two points can be determined by dividing the difference of y-values by the difference of x-values:

The slope of this equation is -4/3. Inputting this into the slope-intercept form of an equation, we get:

To find b, substitute x and y for one of the given coordinate pairs:
0 = (-4/3)(2) + b
0 = -8/3 + b
8/3 = b
Substitute the b value into the equation to finish the line:

Multiply each of the numbers by a ten to a power that they both become whole numbers.
0.67 *

= 67
0.3 *

= 30
Divide as usual.
67 / 30 = 2

So, the answer is 2 1/3.
<span>P(at least 1 ) = 1 - P(exactly none) = 1 - (4/5)^6 = .738
Hope this helps!!!:)</span>
Answer:
The answer is: 81%
Step-by-step explanation:
Divide 405 by 500 and multiply result by 100 to get percent.
405 / 500 = 0.81
0.81 x 100 = 81%
Hope this Helps!! Have an Awesome Day!!
Answer:
Step-by-step explanation:
Hello!
X: number of absences per tutorial per student over the past 5 years(percentage)
X≈N(μ;σ²)
You have to construct a 90% to estimate the population mean of the percentage of absences per tutorial of the students over the past 5 years.
The formula for the CI is:
X[bar] ±
* 
⇒ The population standard deviation is unknown and since the distribution is approximate, I'll use the estimation of the standard deviation in place of the population parameter.
Number of Absences 13.9 16.4 12.3 13.2 8.4 4.4 10.3 8.8 4.8 10.9 15.9 9.7 4.5 11.5 5.7 10.8 9.7 8.2 10.3 12.2 10.6 16.2 15.2 1.7 11.7 11.9 10.0 12.4
X[bar]= 10.41
S= 3.71

[10.41±1.645*
]
[9.26; 11.56]
Using a confidence level of 90% you'd expect that the interval [9.26; 11.56]% contains the value of the population mean of the percentage of absences per tutorial of the students over the past 5 years.
I hope this helps!