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makkiz [27]
2 years ago
8

Nick and Adam workout together. Nick weighs 160 pounds and is gaining about 3 pounds per week. Adam weighs 195 pounds and is los

ing about 2 pounds each week. Write an equation that can be used to find the number of weeks that it will take them to weigh the same amount.
a. Equation:
b. Solution:
c. How much will Nick and Adam weigh?:
Mathematics
1 answer:
kobusy [5.1K]2 years ago
6 0

Answer: 160+(W x 3)=195-(W x 2)

Step-by-step explanation:

I'm not 100% sure this is right but it should be!

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stepan [7]
\bf \begin{array}{ccll}
x&y\\
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if you have already covered slopes, you could also get it that way, in fact is  simpler that way.
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3 years ago
Please help me with the correct answer !!!
Nataly [62]
The answer would be D
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3 years ago
A researcher wishes to estimate the number of households with two cars. How large a sample is needed in order to be 99% confiden
solmaris [256]

Answer:

Sample size n = 1382

so correct option is D) 1382

Step-by-step explanation:

given data

confidence level = 99 %

margin of error = 3%

probability = 25 %

to find out

How large a sample size needed

solution

we know here P = 25 %

so 1 - P = 1 - 0.25

1 - P  = 0.75

and we know E margin of error is 0.03 so value of Z for 99%

α = 1 - 99%   = 1 - 0.99

α  = 0.01

and  \frac{\alpha}{2} = \frac{0.01}{2}

\frac{\alpha}{2}  = 0.005

so Z is here

Z_(\frac{\alpha}{2}) = 2.576

so

sample size will be

Sample size n =  (\frac{(Z_(\frac{\alpha}{2})}{E})^2 * P * (1-P)

put here value

Sample size n = (\frac{2.576}{0.03})^2 * 0.25 * 0.75

Sample size n = 1382

so correct option is D) 1382

3 0
3 years ago
Solve the initial value problems:<br> 1/θ(dy/dθ) = ysinθ/(y^2 + 1); subject to y(pi) = 1
ladessa [460]

Answer:

-\theta cos\thsta+sin\theta = \frac{y^{2} }{2} + ln y + \pi  - \frac{1}{2}

Step-by-step explanation:

Given the initial value problem \frac{1}{\theta}(\frac{dy}{d\theta} ) =\frac{ ysin\theta}{y^{2}+1 } \\ subject to y(π) = 1. To solve this we will use the variable separable method.

Step 1: Separate the variables;

\frac{1}{\theta}(\frac{dy}{d\theta} ) =\frac{ ysin\theta}{y^{2}+1 } \\\frac{1}{\theta}(\frac{dy}{sin\theta d\theta} ) =\frac{ y}{y^{2}+1 } \\\frac{1}{\theta}(\frac{1}{sin\theta d\theta} ) = \frac{ y}{dy(y^{2}+1 )} \\\\\theta sin\theta d\theta = \frac{ (y^{2}+1)dy}{y} \\integrating\ both \ sides\\\int\limits \theta sin\theta d\theta =\int\limits  \frac{ (y^{2}+1)dy}{y} \\-\theta cos\theta - \int\limits (-cos\theta)d\theta = \int\limits ydy + \int\limits \frac{dy}{y}

-\theta cos\thsta+sin\theta = \frac{y^{2} }{2} + ln y +C\\Given \ the\ condition\ y(\pi ) = 1\\-\pi cos\pi +sin\pi  = \frac{1^{2} }{2} + ln 1 +C\\\\\pi + 0 = \frac{1}{2}+ C \\C = \pi  - \frac{1}{2}

The solution to the initial value problem will be;

-\theta cos\thsta+sin\theta = \frac{y^{2} }{2} + ln y + \pi  - \frac{1}{2}

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sergeinik [125]

Answer:

gradient= 1/3 length= sq root 40

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