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3 years ago

2.A cell phone company charges $50 for the phone plus a monthly service charge of $30. The

Mathematics

1 answer:

Mama L [17]3 years ago

8 0

Answer:

o find total, we'll need to add all the costs! There's the cost for the phone which is paid once, monthly fee for all the months you use it.

total = 50 + (30 × months )

In this case, the question wants the equation in x and y, so we can plug those in:

y = 50 + 30x

Step-by-step explanation:

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A rumor spreads through a small town. Let y ( t ) be the fraction of the population that has heard the rumor at time t and assum

Ivan

Answer:

Differential equation

$\frac{dy}{dt} =ky(1-y)$

Solution

$y=\frac{1}{1+4e^{-0.327t}}$

Value of constant k=0.327 days^(-1)

The rumor reaches 80% at 8.48 days.

Step-by-step explanation:

We know

y(t): proportion of people that heard the rumor

y'(t)=ky(1-y), rate of spread of the rumor

Differential equation

$\frac{dy}{dt} =ky(1-y)$

Solving the differential equation

$\frac{dy}{y(1-y)}=k\cdot dt \\\\\int \frac{dx}{y(1-y)} =k \int dt \\\\-ln(1-\frac{1}{y} )+C_0=kt\\\\1-\frac{1}{y} =Ce^{-kt}\\\\\frac{1}{y} =1-Ce^{-kt}\\\\y=\frac{1}{1-Ce^{-kt}}$

Initial conditions:

$y(0)=0.2\\y(3)=0.4\\\\y(0)=0.2=\frac{1}{1-Ce^0}\\\\1-C=1/0.2\\\\C=1-1/0.2= -4\\\\\\y(3)=0.4=\frac{1}{1+4e^{-3k}} \\\\1+4e^{-3k}=1/0.4\\\\e^{-3k}=(2.5-1)/4=0.375\\\\k=ln(0.375)/(-3)=0.327\\\\\\y=\frac{1}{1+4e^{-0.327t}}$

Value of constant k=0.327 days^(-1)

At what time the rumor reaches 80%?

$y(t)=0.8=\frac{1}{1+4e^{-0.327t}} \\\\1+4e^{-0.327t}=1/0.8=1.25\\\\e^{-0.327t}=(1.25-1)/4=0.0625\\\\t=ln(0.0625)/(-0.327)=8.48$

The rumor reaches 80% at 8.48 days.

8 0

3 years ago

Select the correct answer. Why is it important to budget some money for entertainment? A. It is easier to stick to a budget if y

IRISSAK [1]

Answer:

Step-by-step explanation:

well some people like to go on a budget and others prefer not to because they rather spend their money money on wants rather than needs so I really hope this helps.

5 0

2 years ago

Consider a single spin of the spinner. Which events are mutually exclusive? Choose all that apply. a)landing on a shaded sector

denis23 [38]

Answer: the correct options are b and d.

Step-by-step explanation:

Let us first define what mutually exclusive events are. If two events are mutually exclusive, it means that they cannot happen at the same time. For example, getting a head and a tail at the same time are mutually exclusive.

Considering a single spin of the spinner, the events that are mutually exclusive are

b)landing on a shaded sector and landing on a 3. This is because 3 is unshaded. You can either land on 3 or an unshaded sector at a time

d)landing on an unshaded sector and landing on a number less than 2. This is because the numbers in the unshaded sectors are greater than 2.

3 0

3 years ago

Write a system of equations and solve.

Scrat [10]

This looks a lot like a problem previously posted and already worked.
brainly.com/question/8829608

7 0

3 years ago

How can i find the answer to 188x31?

eduard

The answer to 188x31 is 5,828

7 0

3 years ago