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3 years ago

5

Abby practiced for 11/12 of an hour and Lauren practiced for 2/3 of an hour. How long did they practice on Wednesday

2 answers:

Solnce55 [7]3 years ago

6 0

2/3 equals 8/12

11/12 plus 8/12 will give you the improper fraction of 19/12.

19/12 also equals
1 7/12.

Olin [163]3 years ago

5 0

You first have to equalize all of the denominator

2/3 will be equal to : 8/12

Total of their practice time would be :

11/12 + 8/12 = 19/12 hours

Hope this helps

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2/5of a number is 80. What is 34of the same number?

Masja [62]

Answer:

150

Step-by-step explanation:

First you have to find what number 80 is 2/5 of, to do this do the reciprocal of 2/5 which would be 5/2. Multiply 80 by 5/2 and you'll get 200, then find what 3/4 of 200. Multiply 200 by 3, then divide by 4 to get the answer of 150.

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A set of 6 art books costs $241.25. A copy of one of the books, bought separately, costs $38.25. About how much less is the cost

kumpel [21]

The cost of the 6 books if you buy the set is $11.75 less.

The given parameters:

<em>Total number of books, n = 6</em>
<em>Cost of the books, c = $241.25</em>
<em>Cost of one book, = $38.25</em>

The total cost of the 6 books when bought separately is calculated as follows;

Cost = 6 x $38.25

Cost = $229.5

The difference in price is calculated as follows;

= $241.25 - $229.5

= $11.75

Thus, the cost of the 6 books if you buy the set is $11.75 less.

Learn more about cost price here: brainly.com/question/19104371

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3 years ago

Which expression is equivalent to (2x2 +3)(x+4)

FrozenT [24]

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I Hope this Helps!

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3 years ago

During optimal conditions, the rate of change of the population of a certain organism is proportional to the population at time

Lana71 [14]

Answer:

The population is of 500 after 10.22 hours.

Step-by-step explanation:

The rate of change of the population of a certain organism is proportional to the population at time t, in hours.

This means that the population can be modeled by the following differential equation:

$\frac{dP}{dt} = Pr$

In which r is the growth rate.

Solving by separation of variables, then integrating both sides, we have that:

$\frac{dP}{P} = r dt$

$\int \frac{dP}{P} = \int r dt$

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Applying the exponential to both sides:

$P(t) = Ke^{rt}$

In which K is the initial population.

At time t = 0 hours, the population is 300.

This means that K = 300. So

$P(t) = 300e^{rt}$

At time t = 24 hours, the population is 1000.

This means that P(24) = 1000. We use this to find the growth rate. So

$P(t) = 300e^{rt}$

$1000 = 300e^{24r}$

$e^{24r} = \frac{1000}{300}$

$e^{24r} = \frac{10}{3}$

$\ln{e^{24r}} = \ln{\frac{10}{3}}$

$24r = \ln{\frac{10}{3}}$

$r = \frac{\ln{\frac{10}{3}}}{24}$

$r = 0.05$

So

$P(t) = 300e^{0.05t}$

At what time t is the population 500?

This is t for which P(t) = 500. So

$P(t) = 300e^{0.05t}$

$500 = 300e^{0.05t}$

$e^{0.05t} = \frac{500}{300}$

$e^{0.05t} = \frac{5}{3}$

$\ln{e^{0.05t}} = \ln{\frac{5}{3}}$

$0.05t = \ln{\frac{5}{3}}$

$t = \frac{\ln{\frac{5}{3}}}{0.05}$

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LUCKY_DIMON [66]

Answer: The answer is (B)

Step-by-step explanation:

It just took the unit test

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3 years ago