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m_a_m_a [10]
2 years ago
15

Find the coordinates where the line x+y=3 and the curve x^2+3y=27 intersect

Mathematics
1 answer:
mart [117]2 years ago
6 0
<h2>Solving System of Equations by Eliminations</h2>

<h3>Answer:</h3>

(6,-3) and (-3,6)

<h3>Step-by-step explanation:</h3>

The xy-coordinates of intersection between the graphs of two equations are just the xy-values that satisfies both equation.

Let's find the coordinates with the given system of equations:

x +y =3

x^2 +3y = 27

First let's rewrite one of the equations to make some eliminations. The easiest way that I can think of is to multiply both sides of x +y =3 by 3 so when we subtract it from x^2 +3y = 27 the y t terms are eliminated.

x +y =3 \\ 3(x +y) = 3(3) \\ 3x +3y = 9

Subtracting \bold{3x +3y = 9} from \bold{x^2 +3y = 27}:

3x +3y -(x^2 +3y) = 9 -27 \\ 3x +3y -x^2 -3y = 9 -27 \\ -x^2 +3x (3 -3)y = 9 -27 \\ -x^2 +3x + 0y = -18

We can disregard the term with 0 as its coefficient so the result is -x^2 +3x = -18.

Now we can solve the value of x with the resulting equation.

-x^2 +3x = -18 \\ x^2 -3x = 18 \\ x^2 -3x +(\frac{3}{2})^2 = 18 +(\frac{3}{2})^2 \\ ( -\frac{3}{2})^2 = 18 +\frac{9}{4} \\ ( -\frac{3}{2})^2 = \frac{72}{4} +\frac{9}{4} \\ ( -\frac{3}{2})^2 = \frac{81}{4} \\ x -\frac{3}{2} = ±\sqrt{\frac{81}{4}}

Solving for the positive square root:

x -\frac{3}{2} = \sqrt{\frac{81}{4}} \\ x -\frac{3}{2} = \frac{9}{2} \\ x = \frac{9}{2} +\frac{3}{2} \\ x = \frac{12}{2} \\ x = 6

Solving for the negative square root:

x -\frac{3}{2} = -\sqrt{\frac{81}{4}} \\ x -\frac{3}{2} = -\frac{9}{2} \\ x = -\frac{9}{2} +\frac{3}{2} \\ x = -\frac{6}{2} \\ x = -3

We have two x-values that satisfy both of the equation. We also have two respective y-values that satisfy both of the equation. This all means that both equations intersect twice.

Let's solve for the corresponding y-values of each of the x-values with x +y =3.

Solving \bold{y} with \bold{x = 6}:

6 +y = 3 \\ y = 3 -6 \\ y = -3

Now we know that both equations intersect at (6,-3).

Solving \bold{y} with \bold{x = -3}:

-3 +y =3 \\ y = 3 +3 \\ y = 6

Now we know that both equations also intersect at (-3,6)

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A study of the number of business lunches that executives in the insurance and banking industries claim as deductible expenses p
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Answer:

t=\frac{9.1-8}{\sqrt{\frac{(1.9)^2}{40}+\frac{(2.1)^2}{50}}}}=2.604  

p_v =2*P(t_{(88)}>2.604)=0.0108

So the p value is a very low value and using any significance level for given \alpha=0.05 always p_v so we can conclude that we have enough evidence to reject the null hypothesis, and there is enough evidence to conclude that the two means are significantly different at 5%

Step-by-step explanation:

Data given and notation

\bar X_{1}=9.1 represent the mean for the sample 1

\bar X_{2}=8 represent the mean for the sample 2

s_{1}=1.9 represent the sample standard deviation for the sample 1

s_{2}=2.1 represent the sample standard deviation for the sample 2

n_{1}=40 sample size for the group 1

n_{2}=50 sample size for the group 2

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean are different , the system of hypothesis would be:

Null hypothesis:\mu_{1} = \mu_{2}

Alternative hypothesis:\mu_{1} \neq \mu_{2}

If we analyze the size for the samples both are higher than 30 and the population deviations are not given, so for this case is better apply a t test to compare means, and the statistic is given by:

t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}} (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the statistic

We can replace in formula (1) the results obtained like this:

t=\frac{9.1-8}{\sqrt{\frac{(1.9)^2}{40}+\frac{(2.1)^2}{50}}}}=2.604  

Statistical decision

The first step is calculate the degrees of freedom, on this case:

df=n_{1}+n_{2}-2=40+50-2=88

Since is a bilateral test the p value would be:

p_v =2*P(t_{(88)}>2.604)=0.0108

So the p value is a very low value and using any significance level for given \alpha=0.05 always p_v so we can conclude that we have enough evidence to reject the null hypothesis, and there is enough evidence to conclude that the two means are significantly different at 5%

3 0
2 years ago
A city police officer using radar checked the speed of 26 cars as they were travelling down a city street and the data is given
pychu [463]

Answer: The distribution of the data set is positively skewed.

<u>Explanation: </u>

In order to see whether the given data set is symmetric, positively skewed or negatively skewed, we will find the mean, median and mode of the given data set.

For symmetric distribution, Mean = Median=Mode

For positively skewed distribution, Mean > Median>Mode

For negatively skewed distribution, Mean < Median

The mean of the given data set is given below:

Mean = \frac{27+23+22+38+43+24+25+23+22+54+31+30+29+48+27+25+29+28+26+33+25+21+23+34+20+23}{26}

                =\frac{753}{26}

                =28.96

Now, the Median is:

To find the median we need to sort the data in ascending order as:

20,21,22,22,23,23,23,23,24,25,25,25,26,27,27,28,29,29,30,31,33,34,38,43,48,54

Median=\left(\frac{N+1}{2} \right)^{th} item

                   =\left(\frac{26+1}{2}\right)^{th} item

                   =13.5^{th} item

                   =26+0.5 \times (27-26)

                   =26.5

\therefore Median = 26.5

The mode is the most frequently occurring observation. Therefore the mode is:

Mode=23

Since the Mean >Median>Mode, therefore the distribution of the given data set is positively skewed.


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