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2 years ago

Solve graphically the system of linear equations: x-y=1 x+3y=9

Mathematics

2 answers:

Anna [14]2 years ago

5 0

X - y = 1
-x - 3y = -9 = -4y = -8 = y = 2
Then use the easiest equation from before.
Use 2 for y.
x -1(2) = 1
x - 2 = 1 = x = 3
Answer: (3, 2) Hope this helps :)

Nookie1986 [14]2 years ago

3 0

To solve an equation graphically, you need to plot the graph on graph paper and then find the points of intersection (points where both lines meet), which is your answer.

You can't solve it using simultaneous equations.

To plot the graph, you need to make a table of values for x, substitute the random values of x into one equation, find the points that the line touches, and draw the line from the table.

Once you are done drawing both lines, just identify the point at which both lines meet and that's your answer.

NOTE: IF ALL THIS IS TOO CONFUSING, GO TO THE WEBSITE CALLED "DESMOS" AND TYPE IN THE EQUATIONS. IT PLOTS THE GRAPH FOR YOU. ALL YOU NEED TO DO IS FIND THE POINT WHERE BOTH LINES MEET FROM THERE

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A box contains 5 red and 5 blue marbles. Two marbles are withdrawn randomly. If they are the same color, then you win $1.10; if

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Answer:

a) The expected value is $\frac{-1}{15}$

b) The variance is $\frac{49}{45}$

Step-by-step explanation:

We can assume that both marbles are withdrawn at the same time. We will define the probability as follows

#events of interest/total number of events.

We have 10 marbles in total. The number of different ways in which we can withdrawn 2 marbles out of 10 is $\binom{10}{2}$ .

Consider the case in which we choose two of the same color. That is, out of 5, we pick 2. The different ways of choosing 2 out of 5 is $\binom{5}{2}$ . Since we have 2 colors, we can either choose 2 of them blue or 2 of the red, so the total number of ways of choosing is just the double.

Consider the case in which we choose one of each color. Then, out of 5 we pick 1. So, the total number of ways in which we pick 1 of each color is $\binom{5}{1}\cdot \binom{5}{1}$ . So, we define the following probabilities.

Probability of winning: $\frac{2\binom{5}{2}}{\binom{10}{2}}= \frac{4}{9}$

Probability of losing $\frac{(\binom{5}{1})^2}{\binom{10}{2}}\frac{5}{9}$

Let X be the expected value of the amount you can win. Then,

E(X) = 1.10*probability of winning - 1 probability of losing = $1.10\cdot \frac{4}{9}-\frac{5}{9}=\frac{-1}{15}$

Consider the expected value of the square of the amount you can win, Then

E(X^2) = (1.10^2)*probability of winning + probability of losing = $1.10^2\cdot \frac{4}{9}+\frac{5}{9}=\frac{82}{75}$

We will use the following formula

$Var(X) = E(X^2)-E(X)^2$

Thus

Var(X) = $\frac{82}{75}-(\frac{-1}{15})^2 = \frac{49}{45}$

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2 years ago

Biologists conducted a study to investigate the flying velocity of mosquitoes both before and after feeding. The following scatt

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Answer:

It represents a mosquito that flew very fast after feeding relative to all other mosquitoes.

Step-by-step explanation:

The point is unusual because the velocity of the mosquito is substantially greater than all other velocities and is probably an outlier.

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2 years ago

Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of

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<h2><u>Answer with explanation</u>:</h2>

As per given , we have

sample size : n= 65

degree of freedom : df=n-1=64

sample mean : $\overline{x}=19.5$

sample standard deviation : s= 5.2

Since , the population standard deviation is not given , so we apply t-test.

Significance level for 90% confidence : $\alpha=1-0.90=0.10$

t-critical value for significance level 0.10 and df = 64 would be :

$t_c=t_{\alpha/2, df}=t_{0.05,\ 64}=1.6690$

Formula for Confidence interval :

$\overline{x}\pm t_c\dfrac{s}{\sqrt{n}}$

Then , 90% confidence intervals for the population mean number of weekly customer contacts for the sales personnel would be :

$19.5\pm(1.669)\dfrac{5.2}{\sqrt{65}}$

$=19.5\pm1.076$

$(19.5-1.076,\ 19.5+1.076)=(18.424,\ 20.576)$

∴ 90% confidence intervals for the population mean number of weekly customer contacts for the sales personnel : (18.424, 20.576)

Significance level for 95% confidence : $\alpha=1-0.95=0.05$

t-critical value for significance level 0.05 and df = 64 would be :

$t_c=t_{\alpha/2, df}=t_{0.05,\ 64}=1.9977$

Then , 95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel would be :

$19.5\pm(1.9977)\dfrac{5.2}{\sqrt{65}}$

$=19.5\pm1.288$

$(19.5-1.288,\ 19.5+1.288)=(18.212,\ 20.788)$

∴ 95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel : (18.212, 20.788)

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Answer:

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