Answer:
The number of students who scored more than 90 points is 750.
Step-by-step explanation:
Quartiles are statistical measures that the divide the data into four groups.
The first quartile (Q₁) indicates that 25% of the observation are less than or equal to Q₁.
The second quartile (Q₂) indicates that 50% of the observation are less than or equal to Q₂.
The third quartile (Q₃) indicates that 75% of the observation are less than or equal to Q₃.
It is provided that the first quartile is at 90 points.
That is, P (X ≤ 90) = 0.25.
The probability that a student scores more than 90 points is:
P (X > 90) = 1 - P (X ≤ 90)
= 1 - 0.25
= 0.75
The number of students who scored more than 90 points is: 1000 × 0.75 = 750.
Answer:
Step-by-step explanation:
If we take out the extra $3, we can group the bills into one each of $5 and $1, for a value of $6. There will be 7 such groups in the remaining $42.
That means there are 7 bills of the $5 denomination, and 3 more than that (10 bills) of the $1 denomination.
There are 7 $5 bills and 10 $1 bills.
_____
If you want to write an equation, it is usually best to let a variable stand for the most-valuable contributor. Here, we can let x represent then number of $5 bills. Then the value of the cash box is ...
5x +(x+3) = 45
6x = 42 . . . . . . . . subtract 3, collect terms
x = 7 . . . . . . . . . . . there are 7 $5 bills
x+3 = 10 . . . . . . . . there are 10 $1 bills
You may notice that this working parallels the verbal description above. (After we subtract $3, x is the number of $6 groups.)
Answer:
7k
Step-by-step explanation:
−k−(−8k) = -k +8k = 7k
I guess sides and angles !
Answer : 0.0129
Step-by-step explanation:
Given : Based on FAA estimates the average age of the fleets of the 10 largest U.S. commercial passenger carriers is 
years and standard deviation is 
years.
Sample size : 
Let X be the random variable that represents the age of fleets.
We assume that the ages of the fleets of the 10 largest U.S. commercial passenger carriers are normally distributed.
For z-score,
For x=14
By using the standard normal distribution table , the probability that the average age of these 40 airplanes is at least 14 years old will be :-
Hence, the required probability = 0.0129
