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2 years ago

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Based on a poll, 66% of Internet users are more careful about personal information when using a public Wi-Fi hotspot. What is

the probability that among 3 randomly selected Internet users, at least one is more careful about personal information when using a public Wi-Fi hotspot? How is the result affected by the additional information that the survey subjects volunteered to respond?

1 answer:

Harlamova29_29 [7]2 years ago

5 0

Answer:

.

Step-by-step explanation:

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Rhoads earned $324.65 delivering newspapers. She promised her sister 0.2 of her earnings for helping her.How much does Rhoads ow

pashok25 [27]

In math problems, "of" usually means "multiplied by."

(owed to sister) = 0.2*(earnings)
.. = 0.2*$324.65
.. = $64.93

Rhoads owes her sister $64.93.

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3 years ago

The function below can be used to calculate the cost of making a long distance call, f(m), which is based on a $2.50 initial cha

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she was on a call for 36 minutes

Step-by-step explanation: 6.82-2.5= 4.32 4.32/.12= 36

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3 years ago

Select all the points that are on the graph of the line:<br> 2x + 4y=20

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B=10

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3 years ago

According to one cosmological theory, there were equal amounts of the two uranium isotopes 235U and 238U at the creation of the

FromTheMoon [43]

Answer:

6 billion years.

Step-by-step explanation:

According to the decay law, the amount of the radioactive substance that decays is proportional to each instant to the amount of substance present. Let $P(t)$ be the amount of $^{235}U$ and $Q(t)$ be the amount of $^{238}U$ after $t$ years.

Then, we obtain two differential equations

$\frac{dP}{dt} = -k_1P \quad \frac{dQ}{dt} = -k_2Q$

where $k_1$ and $k_2$ are proportionality constants and the minus signs denotes decay.

Rearranging terms in the equations gives

$\frac{dP}{P} = -k_1dt \quad \frac{dQ}{Q} = -k_2dt$

Now, the variables are separated, $P$ and $Q$ appear only on the left, and $t$ appears only on the right, so that we can integrate both sides.

$\int \frac{dP}{P} = -k_1 \int dt \quad \int \frac{dQ}{Q} = -k_2\int dt$

which yields

$\ln |P| = -k_1t + c_1 \quad \ln |Q| = -k_2t + c_2$ ,

where $c_1$ and $c_2$ are constants of integration.

By taking exponents, we obtain

$e^{\ln |P|} = e^{-k_1t + c_1} \quad e^{\ln |Q|} = e^{-k_12t + c_2}$

Hence,

$P = C_1e^{-k_1t} \quad Q = C_2e^{-k_2t}$ ,

where $C_1 := \pm e^{c_1}$ and $C_2 := \pm e^{c_2}$ .

Since the amounts of the uranium isotopes were the same initially, we obtain the initial condition

$P(0) = Q(0) = C$

Substituting 0 for $P$ in the general solution gives

$C = P(0) = C_1 e^0 \implies C= C_1$

Similarly, we obtain $C = C_2$ and

$P = Ce^{-k_1t} \quad Q = Ce^{-k_2t}$

The relation between the decay constant $k$ and the half-life is given by

$\tau = \frac{\ln 2}{k}$

We can use this fact to determine the numeric values of the decay constants $k_1$ and $k_2$ . Thus,

$4.51 \times 10^9 = \frac{\ln 2}{k_1} \implies k_1 = \frac{\ln 2}{4.51 \times 10^9}$

and

$7.10 \times 10^8 = \frac{\ln 2}{k_2} \implies k_2 = \frac{\ln 2}{7.10 \times 10^8}$

Therefore,

$P = Ce^{-\frac{\ln 2}{4.51 \times 10^9}t} \quad Q = Ce^{-k_2 = \frac{\ln 2}{7.10 \times 10^8}t}$

We have that

$\frac{P(t)}{Q(t)} = 137.7$

Hence,

$\frac{Ce^{-\frac{\ln 2}{4.51 \times 10^9}t} }{Ce^{-k_2 = \frac{\ln 2}{7.10 \times 10^8}t}} = 137.7$

Solving for $t$ yields $t \approx 6 \times 10^9$ , which means that the age of the universe is about 6 billion years.

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3 years ago

Peter is reading a 193-page book. He has read three pages more than one fourth of the number of pages he hasnt yet read. How man

Anit [1.1K]

<span>Let n = the number of pages not read
Then the number of pages read = n/4 + 3
Since the total number of pages is 193, we have

n + n/4 + 3 = 193
5n/4 = 190
n = 152
At 8 pages per day it will take 152 pages/ 8 pages/day = 19 days

</span>

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3 years ago