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dsp73
3 years ago
14

the price for an adult movie ticket it 1 1/3 more than a movie ticket for a child. Ines takes her daughter to the movie, buys a

box of popcorn for $5.50 and spends $26.50 write and solve an equation to find prices for each of their movie tickets
Mathematics
2 answers:
dsp733 years ago
6 0
The price of a child ticket is $9.84 and an adult ticket is $11.17
antiseptic1488 [7]3 years ago
5 0

Answer:

The cost of child ticket is $9.3.

The cost of adult ticket is $10.6.

Step-by-step explanation:

Given : The price for an adult movie ticket it 1\frac{1}{3} more than a movie ticket for a child. Ines takes her daughter to the movie, buys a box of popcorn for $5.50 and spends $26.50.

To find : Write and solve an equation to find prices for each of their movie tickets ?

Solution :

Let the price of one child ticket be 'x'.

The price for an adult movie ticket it 1\frac{1}{3}=\frac{4}{3} more than a movie ticket for a child.

i.e. The price for adult ticket is x+\frac{4}{3}

She buys a box of popcorn for $5.50.

Total cost of tickets and popcorn is x+\frac{4}{3}+x+5.50

Total she spends is $26.50.

So, x+\frac{4}{3}+x+5.50=25.50

2x=25.50-5.50-\frac{4}{3}

2x=20-\frac{4}{3}

2x=\frac{60-4}{3}

2x=\frac{56}{3}

x=\frac{56}{3\times 2}

x=\frac{28}{3}

x=9.3

The cost of child ticket is $9.3.

The cost of adult ticket is x+\frac{4}{3}=\frac{28}{3}+\frac{4}{3}=\frac{32}{3}=10.6

The cost of adult ticket is $10.6.

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The answer is the first option
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Find the time required for an investment of 5000 dollars to grow to 6000 dollars at an interest rate of 7.5 percent per year, co
sineoko [7]

2.5 years required for an investment of 5000 dollars to grow to 6000 dollars at an interest rate of 7.5 percent per year, compounded quarterly.

Step-by-step explanation:

The given is,

               Initial investment - $5000

               Future amount - $6000

               Interest rate - 7.5% (compounded quarterly)

Step:1

           Formula to calculate the Future amount with compound interest,

                                   F = P(1+\frac{r}{n} )^{nt}...................................(1)

          Where, F - Future amount

                           P - Initial amount

                            r - Rate of interest

                           n - No. of compounding in a year

                            t - Time period

          From given,

                      F = $6000

                      P = $5000

                       r = 7.5%

                       n = 4 (compounded quarterly)

        Equation (1) becomes,

                              6000=5000(1+\frac{0.075}{4} )^{(t)(4)}

                               \frac{6000}{5000} =(1+0.01875)^{4t}

                                 1.2 = (1.01875)^{4t}

         Take log on both sides,

                            log 1.2 = 4(t) log 1.01875

         Substitute log values,

                       0.07918 = 4(t) (0.0080676)

                                      = (t) (0.0322705)

                                   t = \frac{0.07918}{0.0322705}

                                      = 2.45

                                   t ≅ 2.5 years

Result:

         2.5 years required for an investment of 5000 dollars to grow to 6000 dollars at an interest rate of 7.5 percent per year, compounded quarterly.

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6/15 + 5/15 = 11/15

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