Answer: 0.8413
Step-by-step explanation:
Given : Henry has collected data to find that the typing speeds for the students in a typing class has a normal distribution.
Mean :
Standard deviation : 
Let x be the random variable that represents the typing speeds for the students.
The z-score :-

For x= 51

Using the standard normal distribution table ,the probability that a randomly selected student has a typing speed of less than 51 words per minute :-

Hence, the probability that a randomly selected student has a typing speed of less than 51 words per minute = 0.8413
The s um is positive because 11/12 is greater than 7/8
Step-by-step explanation:
You use the I = PRN formula
P = 11 500
R = 7.25% but you have to 7.25 divide by 100 so it becomes a decimal so it should be 0.0725
N = 2 years and 6 months. This can be 2.5 years because half of 12 is 6
the solution is 11 500 × 0.0725 × 2.5 = 2084.38


Kelly can type 200 words in five minutes.
5y = 4x + 20
- 4x + 5y = 20