Please help this is due today
2 answers:
You might be interested in
Answer:
Per ounce better buy is <em>Happy popcorn</em>.
Step-by-step explanation:
Given that:
Happy popcorn price for 16 ounces = $1.39
Popper popcorn price for 34 ounces = $2.79
Discount coupon present with Gabe = 40 ¢ = $0.40
To find:
Which brand is the better buy per ounce?
Solution:
First of all, let us calculate the price that Gabe has to pay after the discount coupon being applied.
Price for 16 ounces of Happy popcorn after discount = $1.29 - $0.40 = $0.99
Price for 1 ounce of Happy popcorn after discount = = $0.062
Price for 34 ounces of Popper popcorn after discount = $2.79 - $0.40 = $2.39
Price for 1 ounce of Popper popcorn after discount = = $0.070
Clearly, per ounce price of Happy popcorn is lesser than that of Popper popcorn.
Therefore per ounce better buy is <em>Happy popcorn</em>.
Answer:
square units
Step-by-step explanation:
The area of a triangle can be calculated by multiplying the triangle's base by its height, and then dividing the result by . As an algebraic expression, this would be
, where
is the triangle's base and
is the triangle's height. In this case, we know that
and
. Therefore, the area of this triangle is
square units. Hope this helps!
Answer:
a) For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution
b) For this case the data is skewed to the left and we can't assume that we have the normality assumption.
c) This last case the histogram is not symmetrical and the data seems to be skewed.
Step-by-step explanation:
For this case we have the following data:
(a)data = c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)
We can use the following R code to get the histogram
> x1<-c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)
> hist(x1,main="Histogram a)")
The result is on the first figure attached.
For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution
(b)data = c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)
> x2<- c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)
> hist(x2,main="Histogram b)")
The result is on the first figure attached.
For this case the data is skewed to the left and we can't assume that we have the normality assumption.
(c)data = c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)
> x3<-c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)
> hist(x3,main="Histogram c)")
The result is on the first figure attached.
This last case the histogram is not symmetrical and the data seems to be skewed.
