Y = yo + Vot - (gt^2)/2
0 = 0 + 31t - 4.9 t^2
0 = -4.9 t^2 + 31t
Answer: option A.
Answer:
Step-by-step explanation:
In March, Your Co. will collect 20% of January's sales, 30% of February's sales, and 50% of March's sales:
.20×50 +.30×40 +.50×60 = 10 +12 +30 = 52
Similarly, in April, collections will be ...
.20×40 + .30×60 + .50×30 = 8 +18 +15 = 41
Solution:
we are given that
A six sided number cube has faces with the numbers 1 through 6 marked on them.
we have been asked to find the probability that a number less than 2 will occur on one toss of the number cube.
Since a number less than 2 is only one and that is "1" and total number of possible outcome is 6.
and as we know that probability is given using the formula

Substitute the values we get

Hence the required probability is 1/6.
Answer:
There are a total of 23 cars with air conditioning and automatic transmission but not power steering
Step-by-step explanation:
Let A be the cars that have Air conditioning, B the cars that have Automatic transmission and C the cars that have pwoer Steering. Lets denote |D| the cardinality of a set D.
Remember that for 2 sets E and F, we have that

Also,
|E| = |E ∩F| + |E∩F^c|
We now alredy the following:
|A| = 89
|B| = 99
|C| = 74

|(A \cup B \cup C)^c| = 24
|A \ (B U C)| = 24 (This is A minus B and C, in other words, cars that only have Air conditioning).
|B \ (AUC)| = 65
|C \ (AUB)| = 26

We want to know |(A∩B) \ C|. Lets calculate it by taking the information given and deducting more things
For example:
99 = |B| = |B ∩ C| + |B∩C^c| = 11 + |B∩C^c|
Therefore, |B∩C^c| = 99-11 = 88
And |A ∩ B ∩ C^c| = |B∩C^c| - |B∩C^c∩A^c| = |B∩C^c| - |B \ (AUC)| = 88-65 = 23.
This means that the amount of cars that have both transmission and air conditioning but now power steering is 23.
The answer is -3.
If we add positive one/+1 to negative four/-4, we subtract +1 from -4, giving us -3.