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Black_prince [1.1K]
3 years ago
5

The number of bus riders was recorded on one route. The data have these values: minimum = 18, lower quartile = 22, median = 26,

upper quartile = 29, and maximum = 37.
Which box plot represents the data?

Mathematics
2 answers:
Ber [7]3 years ago
4 0

Answer: A

Step-by-step explanation: Because I got it wrong and it said it said it was A

garri49 [273]3 years ago
3 0

Answer:

A

Step-by-step explanation:

I got it right on the assignment and I hope this helps

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How do I put y=90-15x in standard form
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The standard form: Ax + By = C


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8 0
3 years ago
There are 120 people on the bus 50% get off at the first stop 20% get off at the second stop. The rest get off on the third stop
padilas [110]

Answer:

36

Step-by-step explanation:

so 50% of 120 is just half of it. so that would be 60. Then 20% of 120 is 24 so subtract that from 60 and you get 36.

5 0
3 years ago
The projected rate of increase in enrollment at a new branch of the UT-system is estimated by E ′ (t) = 12000(t + 9)−3/2 where E
nexus9112 [7]

Answer:

The projected enrollment is \lim_{t \to \infty} E(t)=10,000

Step-by-step explanation:

Consider the provided projected rate.

E'(t) = 12000(t + 9)^{\frac{-3}{2}}

Integrate the above function.

E(t) =\int 12000(t + 9)^{\frac{-3}{2}}dt

E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+c

The initial enrollment is 2000, that means at t=0 the value of E(t)=2000.

2000=-\frac{24000}{\left(0+9\right)^{\frac{1}{2}}}+c

2000=-\frac{24000}{3}+c

2000=-8000+c

c=10,000

Therefore, E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000

Now we need to find \lim_{t \to \infty} E(t)

\lim_{t \to \infty} E(t)=-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000

\lim_{t \to \infty} E(t)=10,000

Hence, the projected enrollment is \lim_{t \to \infty} E(t)=10,000

8 0
3 years ago
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