The <u>different</u> selections of<u> one</u> meat and <u>one</u> vegetable are possible are 18 selections.
Since the Hickory Stick has a selection of 3 meats and 6 vegetables, the number of ways we can select <u>one </u>meat out of 3 is ³C₁ = 3.
Also, the number of ways we can select <u>one </u>vegetable out of 6 is ⁶C₁ = 6.
So, the total number of selections of <u>one</u> meat and <u>one</u> vegetable is ³C₁ × ⁶C₁ = 3 × 6
= 18 selections
So, the <u>different</u> selections of<u> one</u> meat and <u>one</u> vegetable are possible are 18 selections.
Learn more about combination here:
brainly.com/question/19341024
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!
Answer:
i think it might be 118.4
Step-by-step explanation:
I didn’t quite understand what you were asking, but the origin is (0,0) so you would plot a dot at the origin being (0,0) then depending on your other numbers, you’d plot them using your Y and X