Answer:
33+12t−21t^2
Step-by-step explanation:
(2t-7)²-(5t-4)²
Use binomial theorem (a−b)^2 = a^2−2ab+b^2 to expand (2t-7)².
4t^2−28t+49−(5t-4)²
Use binomial theorem (a−b)^2 = a^2−2ab+b^2 to expand (5t-4)².
4t^2−28t+49−(25t^2−40t+16)
To find the opposite of 25t^2
−40t+16, find the opposite of each term.
4t^2−28t+49−25t^2−40t+16
Combine 4t^2 and −25t^2 to get −21t^2.
−21t^2−28t+49+40t−16
Combine −28t and 40t to get 12t.
−21t^2+12t+49−16
Subtract 16 from 49 to get 33.
−21t^2+12t+33
Swap terms to the left side.
33+12t−21t^2
I hope this helped!
Answer:
The minimum score required for an A grade is 83.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the z-score of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Mean of 72.3 and a standard deviation of 8.
This means that 
Find the minimum score required for an A grade.
This is the 100 - 9 = 91th percentile, which is X when Z has a pvalue of 0.91, so X when Z = 1.34.




The minimum score required for an A grade is 83.
Answer:
A. The tree starts at 0 and increases by 1 every year. The height starts at 5 and increases by 1 every year.
Step-by-step explanation: 5 for the first year, 6 for the second year, 7 for the third year etc.
Answer:
if there can be no more than 5 students, then anything less than 5 students will suffice. However we don’t want any left over ones cause we don’t want that one kid sitting alone during lunch so we can simply put <u><em>4 into each group.</em></u>
Step-by-step explanation:
Hope this helped :DDD
10 n = C
Finding the total would simply be multiplying the cost of the tickets by number of students.