The result is 10/117. We have three distinct letters and four unique non-zero digits. For letters, we have 26 options from A to Z. For digits, there are 9 choices from 1 to 9. As all selections must be distinct, we find the total number of codes as 26 × 25 × 24 × 9 × 8 × 7 × 6. For the specified code, we focus on the combinations with 5 vowels and 4 even digits, leading to a calculation of 5 × 25 × 24 × 8 × 7 × 6 × 4. Probability can thus be expressed as the ratio of these outcomes, yielding: 5 × 25 × 24 × 8 × 7 × 6 × 4 divided by 26 × 25 × 24 × 9 × 8 × 7 × 6.
Answer:
Domain = [0, 50]
Range = [0, 3250]
Step-by-step explanation:
A function illustrates the relationship between two variables (independent and dependent). The independent variable depends on nothing else, serving as the function's input, while the dependent variable relies on the independent variable, acting as the output.
The domain of a function includes all possible input variables (independent variable), and the range encompasses all potential output variables (dependent variables).
For the function C(p) = 65p, p represents the independent variable and C(p) constitutes the dependent variable.
As the hall accommodates a maximum of 50 individuals, the domain of this function is defined as [0, 50]
C(0) results in 65(0) = 0 and C(50) results in 65(50) = 3250
Thus, the range of this function is [0, 3250]
1.4×5=7
0.8×10=8
1.4×10=14
1×15=15
15+14+8+7=44
44÷4=11
LQ of 44=11
LQ=10 minutes
11×3=33
UQ= 29 minutes
The Range is 19 minutes
Detailed breakdown:
Commence with the individual boxes. For determining the number of students in each category, calculate Frequency density × The difference in the category. (if it's 5-15, the difference is 10)
This results in the counts of students in each range.
Next, determine the LQ of 44, which is 11.
Then locate the 11th student's score; in this instance, it resides in the 5-15 range. 7 students have already surpassed it, with 8 in the 5-15 range. Hence, the 11th lies within the bounds of 5-15, making the middle 10.
Repeat this process for the UQ.
The interquartile range is calculated as UQ-LQ, yielding 29-10=19 minutes.
I hope this helps, though I'm not entirely sure if my explanation is coherent and I'm unclear on the terminology I've used for these categories.
The pot has the capability to hold 1/3 more than its current level. Therefore, 1/3 of 5 1/2 quarts equals:
(1/3)(11/2) = 11/6 quarts. Thus, the total capacity of the pot is 11/6 quarts.
Given that the relationship is linear.
The equation can be expressed as y = mx + c.
Substituting, when x = 0, we have y = 32.
Thus, 32 = c.......( 1 )
Then, when x = 100, y results in 212.
Which gives us:
212 = 100m + c.......( 2 )
By equating equations 1 and 2, we obtain:
100m = 212 - 32
Solving for x yields 1.8.
The final equation is therefore y = 1.8x + 32.
Hence, this represents the required solution.
