Answer:
We can use seven letters and numbers.
I am assuming that any numeral in the range 0..9 or any letter from the English alphabet A..Z can appear in any position, with no blank spaces allowed and no restrictions on repetition. I am also assuming that plates with fewer than seven letters and numbers are not allowed.
So, for example A879BX8 is acceptable, so are 5555555 and ABCDEFG, but not A.123.ZX or…..7A, where the dot represents a space.
I am also assuming that you can only use upper case letters.
With these restrictions, there are 36 possibilities for each space and the total number of valid number plates would be 36^7 = 78,364,164,096, let's say about 78 billion.
It is estimated that there are about 1.3 billion cars, trucks and buses in the road today. This number plate system therefore allows more than enough unique license plates. I'd even hazard a guess that it might be more than enough for every road vehicle that has ever been built or ever will be.
In practice there would be other restrictions, for example only letters in some positions and only numbers in others. There'd still be plenty to go around.
Step-by-step explanation:
I multiplied, I converted the percentage to a decimal .42
Your answer should be 9.45.
V=(7*6)/2 *8 the third option
you multiply the height with the base of the triangle and you get the area of the base of the prism so you multiply it with the height of the prism to get the volume
Answer:
B. No, this distribution does not appear to be normal
Step-by-step explanation:
Hello!
To observe what shape the data takes, it is best to make a graph. For me, the best type of graph is a histogram.
The first step to take is to calculate the classmark`for each of the given temperature intervals. Each class mark will be the midpoint of each bar.
As you can see in the graphic (2nd attachment) there are no values of frequency for the interval [40-44] and the rest of the data show asymmetry skewed to the left. Just because one of the intervals doesn't have an observed frequency is enough to say that these values do not meet the requirements to have a normal distribution.
The answer is B.
I hope it helps!
