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Mathematics

1 answer:

aleksklad [387]3 years ago

8 0

Answer:

same here

Step-by-step explanation:

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In a random sample of 30 sixth graders, 9 had traveled outside of the United States. There are a total of 120 sixth graders in t

Aleksandr-060686 [28]

36 students have traveled outside of the United States, when there are total 120 sixth graders.

Explanation

The number of students traveled outside of the United States is 9 when there is total 30 sixth graders. So, the ratio of the student traveled to the total students will be $9: 30 = 3: 10$

Now there are total of 120 sixth graders in the school and lets assume the number of students traveled outside the United states is $x$ . That means the ratio...

$x: 120 = 3:10 \\ \\ \frac{x}{120}=\frac{3}{10} \\ \\ 10x=360\\ \\ x=\frac{360}{10} \\ \\ x= 36$

So, 36 students have traveled outside of the United States.

3 0

3 years ago

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When can you use proportional reasoning to solve a problem? A: When two quantities add to 1 B: When one quantity is a constant m

marshall27 [118]

Answer:

The correct answer will be (B

Step-by-step explanation:

6 0

3 years ago

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Find the solution to the differential equation dB/dt+4B=20 with B(1)=30

natita [175]

Answer:

The solution of the differential equation is $B=5+25e^{-4t+4}$

Step-by-step explanation:

The differential equation $\frac{dB}{dt}+4B=20$ is a first order separable ordinary differential equation (ODE). We know this because a separable first-order ODE has the form:

$y'(t)=g(t)\cdot h(y)$

where g(t) and h(y) are given functions.

We can rewrite our differential equation in the form of a first-order separable ODE in this way:

$\frac{dB}{dt}+4B=20\\\frac{dB}{dt}=20-4B\\\frac{dB}{dt}=4(5-B)\\\frac{1}{5-B}\frac{dB}{dt}=4$

Integrating both sides

$\frac{1}{5-B}\frac{dB}{dt}=4\\\frac{1}{5-B}\cdot dB=4\cdot dt\\\\\int\limits {\frac{1}{5-B}} \, dB=\int\limits {4} \, dt$

The integral of left-side is:

$\int\limits {\frac{1}{5-B}} \, dB\\\mathrm{Apply\:u-substitution:}\:u=5-B\\\int\limits {\frac{1}{5-B}} \, dB=\int\limits {\frac{1}{u}} \, dB\\\mathrm{du=-dB}\\-\int\limits {\frac{1}{u}} \,du\\\mathrm{Use\:the\:common\:integral}:\quad \int \frac{1}{u}du=\ln \left(\left|u\right|\right)\\-\int\limits {\frac{1}{u}} \,du =-\ln \left|u\right|\\\mathrm{Substitute\:back}\:u=5-B\\-\ln \left|5-B\right|\\\mathrm{Add\:a\:constant\:to\:the\:solution}\\-\ln \left|5-B\right|+C$