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Levart [38]
3 years ago
6

Need help #1. The answer is shown, but I don’t know how to get to the answer. Please teach and show steps.

Mathematics
2 answers:
Ivanshal [37]3 years ago
5 0

Answer:

Let x and y be functions of time t such that the sum of x and y is constant.

<h3>(B) is the right answer</h3>

Rufina [12.5K]3 years ago
3 0

Answer:

B

Step-by-step explanation:

We are given that <em>x</em> and <em>y</em> are functions of time <em>t</em> such that <em>x</em> and <em>y</em> is a constant. So, we can write the following equation:

x(t)+y(t)=k,\text{ where $k$ is some constant}

The rate of change of <em>x</em> and the rate of change of <em>y</em> with respect to time <em>t</em> is simply dx/dt and dy/dt, respectively. So, we will differentiate both sides with respect to <em>t: </em>

<em />\displaystyle \frac{d}{dt}\Big[x(t)+y(t)\Big]=\frac{d}{dt}[k]<em />

Remember that the derivative of a constant is always 0. Therefore:

\displaystyle \frac{dx}{dt}+\frac{dy}{dt}=0

And by subtracting dy/dt from both sides, we acquire:

\displaystyle \frac{dx}{dt}=-\frac{dy}{dt}

Hence, our answer is B.

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In a class of 19 students, 3 are math majors. A group of four students is chosen at random. (Round your answers to four decimal
KatRina [158]

Answer:

(a) The probability is 0.4696

(b) The probability is 0.5304

(c) The probability is 0.0929

Step-by-step explanation:

The total number of ways in which we can select k elements from a group n elements is calculate as:

nCx=\frac{n!}{x!(n-x)!}

So, the number of ways in which we can select four students from a group of 19 students is:

19C4=\frac{19!}{4!(19-4)!}=3,876

On the other hand, the number of ways in which we can select four students with no math majors is:

(16C4)*(3C0)=(\frac{16!}{4!(16-4)!})*(\frac{3!}{0!(3-0)!})=1820

Because, we are going to select 4 students form the 16 students that aren't math majors and select 0 students from the 3 students that are majors.

At the same way, the number of ways in which we can select four students with one, two and three math majors are 1680, 360 and 16 respectively, and they are calculated as:

(16C3)*(3C1)=(\frac{16!}{3!(16-3)!})*(\frac{3!}{1!(3-1)!})=1680

(16C2)*(3C2)=(\frac{16!}{2!(16-2)!})*(\frac{3!}{2!(3-1)!})=360

(16C1)*(3C3)=(\frac{16!}{1!(16-1)!})*(\frac{3!}{3!(3-3)!})=16

Then, the probability that the group has no math majors is:

P=\frac{1820}{3876} =0.4696

The probability that the group has at least one math major is:

P=\frac{1680+360+16}{3876} =0.5304

The probability that the group has exactly two math majors is:

P=\frac{360}{3876} =0.0929

6 0
3 years ago
Find my interest if I invest $450 at 8% for 3 years compounded semi-annually.
WINSTONCH [101]

Answer:

The final balance is $571.61.

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Step-by-step explanation:

hope this helps!

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Order of operations<br> 11 + 14 ÷ 2 - 1
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Answer:

17

Step-by-step explanation:

11+14÷2-1

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Answer:

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4 years ago
50 points! help, quick please!
kondor19780726 [428]

Answer:

Step-by-step explanation:

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