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sweet-ann [11.9K]
2 years ago
8

Four people—Rob, Sonja, Jack, and Ang—enter their names into a drawing. The winner receives either a t-shirt or a mug, and which

prize they receive is randomly selected.
What is the probability that either Ang wins and is given a mug, or Jack wins (and is given either prize)? Give the answer as a percent.
Mathematics
2 answers:
weqwewe [10]2 years ago
6 0

Answer:

The probability is 37.5%

Step-by-step explanation:

There are eight total outcomes, but we only need to focus on three. Ang winning a mug and Jack winning either a t-shirt or a mug, that makes three. Write that as a fraction: 3/8, and then divide. 3 divided by 8 gives us 0.375. But we need a percent so multiply that by 100, that now gives us 37.5.

So, the probability that either Ang wins and is given a mug, or Jack wins is 37.5%. If it makes you feel more confident in this, I put the same exact answer for my assessment and I got it correct.

Also, “The monks named me aOng.”

Snowcat [4.5K]2 years ago
3 0

Answer: 50%

Step-by-step explanation: Good luck! :D

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3 years ago
A large corporation starts at time t = 0 to invest part of its receipts continuously at a rate of P dollars per year in a fund f
Andrews [41]

Answer:

A = \frac{P}{r}\left( e^{rt} -1 \right)

Step-by-step explanation:

This is <em>a separable differential equation</em>. Rearranging terms in the equation gives

                                                \frac{dA}{rA+P} = dt

Integration on both sides gives

                                            \int \frac{dA}{rA+P} = \int  dt

where c is a constant of integration.

The steps for solving the integral on the right hand side are presented below.

                               \int \frac{dA}{rA+P} = \begin{vmatrix} rA+P = m \implies rdA = dm\end{vmatrix} \\\\\phantom{\int \frac{dA}{rA+P} } = \int \frac{1}{m} \frac{1}{r} \, dm \\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \int \frac{1}{m} \, dm\\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |m| + c \\\\&\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |rA+P| +c

Therefore,

                                        \frac{1}{r} \ln |rA+P| = t+c

Multiply both sides by r.

                               \ln |rA+P| = rt+c_1, \quad c_1 := rc

By taking exponents, we obtain

      e^{\ln |rA+P|} = e^{rt+c_1} \implies  |rA+P| = e^{rt} \cdot e^{c_1} rA+P = Ce^{rt}, \quad C:= \pm e^{c_1}

Isolate A.

                 rA+P = Ce^{rt} \implies rA = Ce^{rt} - P \implies A = \frac{C}{r}e^{rt} - \frac{P}{r}

Since A = 0  when t=0, we obtain an initial condition A(0) = 0.

We can use it to find the numeric value of the constant c.

Substituting 0 for A and t in the equation gives

                         0 = \frac{C}{r}e^{0} - \frac{P}{r} \implies \frac{P}{r} = \frac{C}{r} \implies C=P

Therefore, the solution of the given differential equation is

                                   A = \frac{P}{r}e^{rt} - \frac{P}{r} = \frac{P}{r}\left( e^{rt} -1 \right)

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Step-by-step explanation:

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Read 2 more answers
how much antifreeze 25% antifreeze and 50% antifreeze should be combined to give 40 gallon of 30% antifreeze?​
VARVARA [1.3K]

32 gallons of 25 % antifreeze should be combined with 8 gallons of 50 % antifreeze to get 40 gallon of 30% antifreeze

<h3><u>Solution:</u></h3>

From given,

Final mixture is 40 gallon

Let "x" be the gallon of 25 % antifreeze

Then, (40 - x) is the gallon of 50 % antifreeze

Therefore, according to question,

"x" gallons of 25 % antifreeze should be combined with (40 - x) gallons of 50 % antifreeze to get 40 gallon of 30% antifreeze

<h3><u>Therefore, we frame a equation as:</u></h3>

25 % of x + 50 % of (40 - x) = 30 % of 40

Solve for "x"

\frac{25}{100} \times x + \frac{50}{100} \times (40-x) = \frac{30}{100} \times 40\\\\0.25x + 0.5(40-x) = 12\\\\0.25x + 20 - 0.5x = 12\\\\0.25x - 0.5x = 12 - 20\\\\-0.25x = -8\\\\0.25x = 8\\\\Divide\ both\ sides\ by\ 0.25\\\\x = 32

Thus, 32 gallons of 25 % antifreeze is used

Then, (40 - x) = (40 - 32) = 8

Thus 8 gallons of 50 % antifreeze is used

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2 years ago
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