Let the school hire x buses and y vans.
A bus can hold 40 students and 3 teachers.
A van can hold 8 students and 1 teacher.
The number of students riding in buses and vans is at least 400, therefore
40x + 8y ≥ 400 (1)
The number of teachers riding in buses and vans is at most 36, therefore
3x + y ≤ 36 (2)
Write (1) and (2) as
y ≥ 50 - 5x (3)
y ≤ 36 - 3x (4)
The equality portion of the solution of (3) and (4) is
36 - 3x = 50 - 5x
2x = 14
x = 7 => y = 36 - 3*7 = 15
A graph of the inequalities indicates the acceptable solution in shaded color, as shown below.
The minimum cost of renting buses and vans is
7*$1200 + 15*$100 = $9900
Answer: The minimum cost is $9,900
Answer:
212.9 years
Step-by-step explanation:
Given that the half-life of cesium-137 is 30 years. Suppose we have a 170 mg sample
P0 = 175
P(30) = 87.5
So we can write equation as

b) After 60 years t = 30
In 30 years it becomes half and hence in 60 years it would become 1/4
i.e.
mg
c) If P(t) =1, let us find t

212.9
Answer:
D. The difference of the means is not significant because the re-randomizations show that it is within the range of what could happen by chance.
Step-by-step explanation:
The treatment group using System A reported a mean of 18.5 lost bags per day. The treatment group using System B reported a mean of 16.6 lost bags per day.
The best conclusion that can be made is - The difference of the means is not significant because the re-randomizations show that it is within the range of what could happen by chance.
As we know, in statistics, nothing happens by chance. So, this option is correct.
Answer:
P(made 2nd attempt|made 1st attempt)=P(made 2nd attempt)
Step-by-step explanation:
Here given that a basketball player that shoots 80% from the free throw line attempts two free throws.
If x is the no of shoots he makes (say) then we find that each throw is independent of the other.
In other words, because he made successful first attempt, his chances for second attempt will not change
Prob for success in each attempt remains the same as 0.80
Hence I throw is independent of II throw.
When A and B are independent,then we have
P(A/B) = P(A)
Hence answer is
P(made 2nd attempt|made 1st attempt)=P(made 2nd attempt)