2x^3y+18xy-10x^2y-90y
= 2xy(x^2 + 9)- 10y(x^2 + 9)
= (x^2 + 9)(2xy - 10y)
= (x^2 + 9) 2y(x - 5)
= 2y (x^2 + 9) (x - 5)
Answer:
It will never cost the same amount.
But, It would take 5 classes for Erik to pass Jason.
Step-by-step explanation:
Jason: 5 classes×$2 per class=$10+$50=60$
Erik: $13×5 classes= $65
Answer:
I think it's 12×54 but im probs wrong
From the histogram it can be seen that
1.) the data is not qualitative. (<span>Qualitative data is typically descriptive <span>data and is not measured by a number but age is measured by a number).
2.) </span></span><span>The data is not skewed right. (</span><span>For a right skewed distribution, the tail of the distribution on the right hand (positive) side is longer than on the left hand side).
3.) </span><span>There were 31 people at the party. (The total number of people at the party from the histogram is given by the sum of the heights of the bars of the histogram = 5 + 13 + 8 + 3 + 2 = 31).
4.) </span><span>There were children at the party. (From the histogram, there were 5 poeple aged 20 and below, which we can describe as children).
5.) </span><span>The most common age of the attendees was not 13 years old. (The mode (i.e. most occuring frequency) of a distribution is represented by the height of the tallest bar of the histogram. From the histogram, the tallest bar represents the ages between 21 to 40)
6.) </span><span>The majority of the people at the party were less than 61 years old. (From the histogram, there were 5 + 13 + 8 = 26 people below the age of 61 in the party and 3 + 2 = 5 people above the age of 60 in the party).
Therefore, the </span><span>statements that must be true according to the histogram are
</span><span>There were 31 people at the party.
</span><span>There were children at the party.
and
</span><span>The majority of the people at the party were less than 61 years old.</span>
