By solving a linear equation, we will see that the total cost for renting the bus is $90.
<h3>What was the total cost of renting the bus, in dollars?</h3>
Let's say that the total cost is C.
When there are 20 students, each student should pay:
p = C/20
When the other 10 students are added (for a total of 30) each student pays:
p' = C/30.
We know that the cost for each of the original 20 students decreased by $1.50, so:
p' = p - $1.50
Then we have 3 equations to work with:
p = C/20
p' = C/30.
p' = p - $1.50
Now we can replace the first and second equations into the third one:
C/30 = C/20 - $1.50
Now we can solve this linear equation for C:
C/20 - C/30 = $1.50
C*( 1/20 - 1/30) = $1.50
C*(30/600 - 20/600) = $1.50
C*(10/600) = $1.50
C*(1/60) = $1.50
C = 60*$1.50 = $90
So the total cost for renting the bus is $90.
If you want to learn more about linear equations:
brainly.com/question/1884491
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Answer:
he should chose compounded annually because he would have and extra 32.12 dollars at the end of the 4 years
Step-by-step explanation:
hope this helps
Answer:
No, the on-time rate of 74% is not correct.
Solution:
As per the question:
Sample size, n = 60
The proportion of the population, P' = 74% = 0.74
q' = 1 - 0.74 = 0.26
We need to find the probability that out of 60 trains, 38 or lesser trains arrive on time.
Now,
The proportion of the given sample, p = 
Therefore, the probability is given by:
![P(p\leq 0.634) = [\frac{p - P'}{\sqrt{\frac{P'q'}{n}}}]\leq [\frac{0.634 - 0.74}{\sqrt{\frac{0.74\times 0.26}{60}}}]](https://tex.z-dn.net/?f=P%28p%5Cleq%200.634%29%20%3D%20%5B%5Cfrac%7Bp%20-%20P%27%7D%7B%5Csqrt%7B%5Cfrac%7BP%27q%27%7D%7Bn%7D%7D%7D%5D%5Cleq%20%5B%5Cfrac%7B0.634%20-%200.74%7D%7B%5Csqrt%7B%5Cfrac%7B0.74%5Ctimes%200.26%7D%7B60%7D%7D%7D%5D)
P![(p\leq 0.634) = P[z\leq -1.87188]](https://tex.z-dn.net/?f=%28p%5Cleq%200.634%29%20%3D%20P%5Bz%5Cleq%20-1.87188%5D)
P![(p\leq 0.634) = P[z\leq -1.87] = 0.0298](https://tex.z-dn.net/?f=%28p%5Cleq%200.634%29%20%3D%20P%5Bz%5Cleq%20-1.87%5D%20%3D%200.0298)
Therefore, Probability of the 38 or lesser trains out of 60 trains to be on time is 0.0298 or 2.98 %
Thus the on-time rate of 74% is incorrect.
