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3 years ago

The maximum capacity in the

Mathematics

2 answers:

Igoryamba3 years ago

5 0

Answer:

34 classes

Step-by-step explanation:

First, subract 180 from 1200. You should get 1020. then do 1020/30. So 34 classes can watch the performance of 180 students.

Arada [10]3 years ago

5 0

Answer:

Step-by-step explanation:

If there are 180 students performing you need to subtract that from 1200, which leaves you with 1020

Then you divide it by 30, since there are 30 people in each class

You should then end up with 34 classes

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nekit [7.7K]

Hey there!

-4.5 * (-22.1)

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Therefore, your answer is:

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3 years ago

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ruslelena [56]

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I need help urgent plz someone help me solved this problem! Can someone plz help I’m giving you 10 points! I need help plz help

VMariaS [17]

Answer: a) 8.779 years

b) 8.664 years

<u>Step-by-step explanation:</u>

$A=P\bigg(1+\dfrac{r}{n}\bigg)^{nt}$

A: accumulated amount (balance)
P: principal amount (original/initial investment)
r: interest rate (convert to a decimal)
n: number of times compounded per year
t: number of years

Given: A = 1800, P = 900, r = 8% = 0.08, n = 3, t = unknown

$1800=900\bigg(1+\dfrac{0.08}{3}\bigg)^{3t}\\\\\\2=\bigg(1+\dfrac{0.08}{3}\bigg)^{3t}\\\\\\ln\ 2=ln \bigg(1+\dfrac{0.08}{3}\bigg)^{3t}\\\\\\ln\ 2=3t\ ln\bigg(1+\dfrac{0.08}{3}\bigg)\\\\\\\dfrac{ln\ 2}{3\ ln\bigg(1+\dfrac{0.08}{3}\bigg)}=t\\\\\\\large\boxed{8.779=t}$

$A=Pe^{rt}$

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