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2 years ago

Which of the following numbers represents the fraction 8/9 to the nearest percent?

Mathematics

1 answer:

jekas [21]2 years ago

6 0

The number that presents 8/9 to the nearest percent is .89.

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The sum of two numbers is 39. The second number is 6 more than half of the first number. Let a represent the first number and le

inn [45]

Write 2 equations:

a +b = 39

and b = 1/2a + 6

Now replace b in the first equation with the second one:

a + 1/2a + 6 = 39

Combine like terms:

1 1/2a + 6 = 39

Subtract 6 from each side:

1 1/2a = 33

Divide both sides by 1 1/2:

a = 33 / 1 1/2 = 33 / 1.5

a = 22

Now replace a with 22 in the first equation and solve for b:

22 + b = 39

b = 39 -22

b = 17

The two numbers are 17 and 22.

8 0

3 years ago

Read 2 more answers

What is the area of a circle with a diameter of 36 millimeters?

Aleonysh [2.5K]

1st step is to find the radius, then substitute into the formula for area of a circle

$A = \pi r^2~~~~~~r = radius$

to find the radius, just halve the diameter

6 0

3 years ago

Read 2 more answers

% of 50 shirts is 35 shirts

Nitella [24]

the answer is 70% your welcome

4 0

3 years ago

Ms Lucas has 9 cases of crayon with 52 boxes in each case Ms Bruns has 6 cases of crayon with a 104 in each case How many boxes

saveliy_v [14]

Answer:

1092

Step-by-step explanation:

Step one:

given data

Ms Lucas has 9 cases of crayon with 52 boxes in each case

total crayon is

9*52= 468

Ms Bruns has 6 cases of crayon with a 104 in each case

total crayon is

6*104= 624

The total crayon is

468+624= 1092

5 0

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:

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

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

$\ln{P} = rt + K$

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

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

$1000 = 300e^{24r}$

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

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

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

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

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

$r = 0.05$

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

At what time t is the population 500?

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

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

$500 = 300e^{0.05t}$

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

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

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

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

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

$t = 10.22$

The population is of 500 after 10.22 hours.

7 0

3 years ago