<span>-Both box plots show the same interquartile range.
>Interquartile range (IQR) is computed by Q3-Q1.
For Mr. Ishimoto's class, Q3 is 35 and Q1 is 31. 35-31 = 4.
For Ms. Castillo's class, Q3 is 34 and Q1 is 30. 34-30 = 4.
</span><span>-Mr. Ishimoto had the class with the greatest number of students.
>Mr. Ishimoto had 40 students, represented by the last data point of the whiskers.
</span><span>-The smallest class size was 24 students.
>Which was Ms. Castillo's class.</span>
Answers:
Row 0: 1
Row 1: 1 1
Row 2: 1 2 1
Row 3: 1 3 3 1
Row 4: 1 4 6 4 1
=========================================
Explanation:
This is known as Pascal's Triangle. Each value, other than the '1's, is found by adding the values directly above it. For instance, the '4' is the result of 1+3, and the 6 is a result of 3+3, and so on.
You can use the nCr combination formula to find the same values.
Alternatively you can expand out binomials of the form (x+y)^n, for integer values of n, and you'll find the coefficients of the terms match perfectly with Pascal's Triangle.
Answer:
4.43
Step-by-step explanation:
7.90
- 3.47
__________
4. 4 3
I couldn't regroup but of you cross out 0 and put 10, 10-7= 3
and you have to borrow one from 9 and make it 8 and 8-4=4
Initial repair cost = $1200
Gratuity = 15% of the initial repair cost
= 15% of 1200
= 15 × 12
= 180
Hence, the gratuity for the service is $180.
$250 c+ $ 180 g > $ 950
<u>Step-by-step explanation:</u>
As a cryptographer (c), Miyoko earns per day = $ 250
As a geologist (g) , Miyoko earns per day = $ 180
So the equation comes to be $250 c+ $ 180 g = $ 950
The equation can be rewritten to find c as, (950-180 g) / 250
The equation can be rewritten to find g as, (950 - 250 c) / 180
Plugin different values of c and g in the above 2 equations, we can find that ,
To achieve the goal, Miyoko requires to be a geologist for 3 days and crpytographist for 2 days.
