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3 years ago

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The points (-3,-1) and (-3,5) are adjacent vertices of a rectangle. Two of the sides of the rectangle have a length of 8 units.

What is the length of a diagonal of the rectangle?

1 answer:

taurus [48]3 years ago

6 0

Answer:

The answer to your question is 10 units

Step-by-step explanation:

Data

A (-3, -1)

B (-3 , 5)

length of a side = 8 units

Process

1.- Calculate the distance between A and B

dAB = $\sqrt{(x2 - x1)^{2} + (y2 - y1)^{2}}$

x1 = -3 y1 = -1

x2 = -3 y2 = 5

-Substitution

dAB = $\sqrt{(-3 + 3)^{2}+ (5 + 1)^{2}}$

dAB = $\sqrt{0^{2} + 6^{2}}$

dAB = $\sqrt{36}$

dAB = 6 u

2.- Calculate the diagonal using the Pythagorean theorem.

c² = a² + b²

c² = 6² + 8²

c² = 36 + 64

c² = 100

c = 10 units

- Conclusion

The diagonal of the rectangle measures 10 units

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Step-by-step explanation:

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4 years ago

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According to one investigator’s model, the data are like 400 draws made at random from a large box. The null hypothesis says tha

tankabanditka [31]

Answer:

$z=\frac{52.75-50}{\frac{25}{\sqrt{400}}}=2.2$

$p_v =2*P(Z>2.2)=0.0278$

If we compare the p value and the significance level assumed $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the mean is significantly diffrent fro 50 at 1% of signficance.

Step-by-step explanation:

1) Data given and notation

$\bar X=52.75$ represent the mean height for the sample

$s=25$ represent the sample standard deviation

$n=400$ sample size

$\mu_o =50$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is equal to 50 or not, the system of hypothesis would be:

Null hypothesis: $\mu = 50$

Alternative hypothesis: $\mu \neq 50$

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value but we can assume it as z distribution, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{52.75-50}{\frac{25}{\sqrt{400}}}=2.2$

P-value

Since is a two sided test the p value would be:

$p_v =2*P(Z>2.2)=0.0278$

Conclusion

If we compare the p value and the significance level assumed $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the mean is significantly diffrent fro 50 at 1% of signficance.

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4 years ago

Solve these linear equations by Cramer's Rules Xj=det Bj / det A:

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Answer:

(a) $x_1=-2,x_2=1$

(b) $x_1=\frac{3}{4} ,x_2=-\frac{1}{2} ,x_3=\frac{1}{4}$

Step-by-step explanation:

(a) For using Cramer's rule you need to find matrix $A$ and the matrix $B_j$ for each variable. The matrix $A$ is formed with the coefficients of the variables in the system. The first step is to accommodate the equations, one under the other, to get $A$ more easily.

$2x_1+5x_2=1\\x_1+4x_2=2$

$\therefore A=\left[\begin{array}{cc}2&5\\1&4\end{array}\right]$

To get $B_1$ , replace in the matrix A the 1st column with the results of the equations:

$B_1=\left[\begin{array}{cc}1&5\\2&4\end{array}\right]$

To get $B_2$ , replace in the matrix A the 2nd column with the results of the equations:

$B_2=\left[\begin{array}{cc}2&1\\1&2\end{array}\right]$

Apply the rule to solve $x_1$ :

$x_1=\frac{det\left(\begin{array}{cc}1&5\\2&4\end{array}\right)}{det\left(\begin{array}{cc}2&5\\1&4\end{array}\right)} =\frac{(1)(4)-(2)(5)}{(2)(4)-(1)(5)} =\frac{4-10}{8-5}=\frac{-6}{3}=-2\\x_1=-2$

In the case of B2, the determinant is going to be zero. Instead of using the rule, substitute the values of the variable $x_1$ in one of the equations and solve for $x_2$ :

$2x_1+5x_2=1\\2(-2)+5x_2=1\\-4+5x_2=1\\5x_2=1+4\\ 5x_2=5\\x_2=1$

(b) In this system, follow the same steps,ust remember $B_3$ is formed by replacing the 3rd column of A with the results of the equations:

$2x_1+x_2 =1\\x_1+2x_2+x_3=0\\x_2+2x_3=0$

$\therefore A=\left[\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right]$

$B_1=\left[\begin{array}{ccc}1&1&0\\0&2&1\\0&1&2\end{array}\right]$

$B_2=\left[\begin{array}{ccc}2&1&0\\1&0&1\\0&0&2\end{array}\right]$

$B_3=\left[\begin{array}{ccc}2&1&1\\1&2&0\\0&1&0\end{array}\right]$

$x_1=\frac{det\left(\begin{array}{ccc}1&1&0\\0&2&1\\0&1&2\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)} =\frac{1(2)(2)+(0)(1)(0)+(0)(1)(1)-(1)(1)(1)-(0)(1)(2)-(0)(2)(0)}{(2)(2)(2)+(1)(1)(0)+(0)(1)(1)-(2)(1)(1)-(1)(1)(2)-(0)(2)(0)}\\ x_1=\frac{4+0+0-1-0-0}{8+0+0-2-2-0} =\frac{3}{4} \\x_1=\frac{3}{4}$

$x_2=\frac{det\left(\begin{array}{ccc}2&1&0\\1&0&1\\0&0&2\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)} =\frac{(2)(0)(2)+(1)(0)(0)+(0)(1)(1)-(2)(0)(1)-(1)(1)(2)-(0)(0)(0)}{4} \\x_2=\frac{0+0+0-0-2-0}{4}=\frac{-2}{4}=-\frac{1}{2}\\x_2=-\frac{1}{2}$

$x_3=\frac{det\left(\begin{array}{ccc}2&1&1\\1&2&0\\0&1&0\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)}=\frac{(2)(2)(0)+(1)(1)(1)+(0)(1)(0)-(2)(1)(0)-(1)(1)(0)-(0)(2)(1)}{4} \\x_3=\frac{0+1+0-0-0-0}{4}=\frac{1}{4}\\x_3=\frac{1}{4}$

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Answer:

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Step-by-step explanation:

Respective angles of both triangles are 40°, 50° and 90°

AA similarity is the case:

<em>If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.</em>

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