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3 years ago

15

There were 27 women and 22 men who stood in line to buy the first tickets for a new movie. By the end of the first hour, 4 times

that number of tickets had been sold the the movie theater’s web site. How many were sold in all

1 answer:

svlad2 [7]3 years ago

8 0

49 × 4= 196 tickets i believe

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Answer:

Principal = $1,000

Rate of interest = 14% = 0.14

Time = 3 years

$A= P(1+r)^t$

A = Amount or future value

P = Principal

r = rate of interest in decimals

t = time

Substitute the values in the formula

$A= 1000(1+0.14)^3$

$A=1481.544$

Future Value is $1481.544

Compound interest = Amount - Principal = $1481.544- $1000 = $481.544

Formula of simple interest : $SI =\frac{P \times R \times T}{100}$

Substitute the values :

$SI =\frac{1000 \times 14 \times 3}{100}$

$SI =420$

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Montano1993 [528]

Answer:

$\dfrac{m^3}{16p^2q^2}$

Step-by-step explanation:

Given:

$(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}$

1.

$m^{-1}=\dfrac{1}{m}$

2.

$q^0=1$

3.

$2pm^{-1}q^0=2p\cdot \dfrac{1}{m}\cdot 1=\dfrac{2p}{m}$

4.

$(2pm^{-1}q^0)^{-4}=\left(\dfrac{2p}{m}\right)^{-4}=\left(\dfrac{m}{2p}\right)^4=\dfrac{m^4}{(2p)^4}=\dfrac{m^4}{16p^4}$

5.

$m^{-1}=\dfrac{1}{m}$

6.

$2m^{-1} p^3=2\cdot \dfrac{1}{m}\cdot p^3=\dfrac{2p^3}{m}$

7.

$\dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{\frac{2p^3}{m}}{2pq^2}=\dfrac{2p^3}{m}\cdot \dfrac{1}{2pq^2}=\dfrac{p^2}{mq^2}$

8.

$(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{m^4}{16p^4}\cdot \dfrac{p^2}{mq^2}=\dfrac{m^3}{16p^2q^2}$

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3 years ago