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4 years ago

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20 POINTS!!! HELP ME!!!! Use the linear combination method to solve the system of equations. Explain each step of your solution.

3x-8y=7" " " " x+2y=-7

1 answer:

madreJ [45]4 years ago

7 0

The solution of the system of equations is (-3 , -2)

Step-by-step explanation:

Steps for Using Linear Combinations Method)

Arrange the equations with like terms in columns
Analyze the coefficients of x or y
Add the equations and solve for the remaining variable
Substitute the value into either equation and solve

∵ 3 x - 8 y = 7 ⇒ (1)

∵ x + 2 y = -7 ⇒ (2)

- Multiply equation (2) by 4 to make the coefficients of y are equal in

magnitude and different in sign

∴ 4 x + 8 y = -28 ⇒ (3)

Add equations (1) and (3)

∵ 3 x - 8 y = 7 ⇒ (1)

∵ 4 x + 8 y = -28 ⇒ (3)

∴ 7 x = -21

- Divide both sides by 7

∴ x = -3

Substitute the value of x in equation (2) to find y

∵ x + 2 y = -7 ⇒ (2)

∵ x = -3

∴ -3 + 2 y = -7

- Add 3 to both sides

∴ 2 y = -4

- Divide both sides by 2

∴ y = -2

The solution of the system of equations is (-3 , -2)

Learn more:

You can learn more about the system of the linear equations in brainly.com/question/13168205

#LearnwithBrainly

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In a survey, the planning value for the population proportion is p* = 0.35. How large a sample should be taken to provide a 95%

lions [1.4K]

Answer:

$n=\frac{0.35(1-0.35)}{(\frac{0.05}{1.96})^2}=349.59$

And rounded up we have that n=350

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$p$ represent the real population proportion of interest

$\aht p$ represent the estimated proportion for the sample

n is the sample size required (variable of interest)

$z$ represent the critical value for the margin of error

Solution to the problem

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.05$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

And replacing into equation (b) the values from part a we got:

$n=\frac{0.35(1-0.35)}{(\frac{0.05}{1.96})^2}=349.59$

And rounded up we have that n=350

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3 years ago

A roast turkey is taken from an oven when its temperature has reached 185 degrees F and is placed on a table in a room where the

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(a) Using Newton's Law of Cooling, $\dfrac{dT}{dt} = k(T - T_s)$ , we have $\dfrac{dT}{dt} = k(T - 75)$ where T is temperature after T minutes.
Separate by dividing both sides by T - 75 to get $\dfrac{dT}{T - 75} = k dt$ . Integrate both sides to get $\ln|T - 75| = kt + C$ .

Since $T(0) = 185$ , we solve for C:
$|185 - 75| = k(0) + C\ \Rightarrow\ C = \ln 110$
So we get $\ln|T - 75| = kt + \ln 110$ . Use T(30) = 150 to solve for k:
$\ln| 150 - 75 | = 30k + \ln 110\ \Rightarrow\ \ln 75 - \ln 110= 30k \Rightarrow \\ k= \frac{1}{30}\ln (75/110) = \frac{1}{30}\ln(15/22)$

So

$\ln|T - 75| = kt + \ln 110 \Rightarrow |T - 75| = e^{kt + \ln110} \Rightarrow \\ \\
|T - 75| = 110e^{kt} \Rightarrow T - 75 = \pm110e^{(1/30)\ln(15/22)t} \Rightarrow \\
T = 75 \pm110e^{(1/30)\ln(15/22)t}$

But choose Positive because T > 75. Temp of turkey can't go under.

$T(t) = 75 + 110e^{(1/30)\ln(15/22)t} \\
T(45) = 75 + 110e^{(1/30)\ln(15/22)(45)} = 136.929 \approx 137{}^{\circ}F$

(b)

$T(t) = 75 + 110e^{(1/30)\ln(15/22)t} = 75 + 110(15/22)^{t/30} \\
100 = 75 + 110(15/22)^{t/30} \\
25 = 110(15/22)^{t/30} 
\frac{25}{110} = (15/22)^{t/30} \\
\ln(25/110) / ln(15/22) = t/30 \\
t = 30\ln(25/110) / ln(15/22) \approx 116\ \mathrm{min}$

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