What is 2 and 2/3 x 5

Select the pair of equations whose graphs are perpendicular.

viktelen [127]

$k:\ y=m_1x+b_1\ \ \ \ and\ \ \ \ l:\ y=m_2x+b_2\\ \\the\ line\k\ is\ perpendicular\ to\ the\ line\ l\ \ \ \Leftrightarrow\ \ \ m_1\cdot m_2=-1\\----------------------------\\\\A.\\2y=-3x+5\ \ \Rightarrow\ \ \ y=- \frac{3}{2} x+2.5\ \ \Rightarrow\ \ m_1=- \frac{3}{2} \\\\2x+3y=4\ \ \Rightarrow\ \ 3y=-2x+4\ \ \Rightarrow\ \ y=- \frac{2}{3}x+1 \frac{1}{3} \ \ \Rightarrow\ \ m_2=- \frac{2}{3} \\\\m_1\cdot m_2=- \frac{3}{2} \cdot (- \frac{2}{3})=1 \neq -1$

$B.\\5x-8y=9\ \ \Rightarrow\ \ -8y=-5x+9\ \ \Rightarrow\ \ \ y= \frac{5}{8} x-1 \frac{1}{8} \ \Rightarrow\ \ m_1= \frac{5}{8} \\\\12x-5y=7\ \Rightarrow\ \ -5y=-12x+7\ \ \Rightarrow\ \ y= \frac{12}{5}x-1 \frac{2}{5} \ \ \Rightarrow\ \ m_2= \frac{12}{5} \\\\m_1\cdot m_2= \frac{5}{8} \cdot \frac{12}{5}= \frac{3}{2} \neq -1$

$C.\\y=2x-7\ \ \Rightarrow\ \ m_1=2 \\\\x+2y=3\ \Rightarrow\ \ 2y=-x+3\ \ \Rightarrow\ \ y= -\frac{1}{2}x+1 \frac{1}{2}\ \ \Rightarrow\ \ m_2=- \frac{1}{2} \\\\m_1\cdot m_2= 2 \cdot (- \frac{1}{2})= -1\ \ \Rightarrow\ \ y=2x-7\ \ \ \ \perp\ \ \ \ x+2y=3$

$
D.\\x+6y=8\ \ \Rightarrow\ \ 6y=-x+8\ \ \Rightarrow\ \ \ y=- \frac{1}{6} x+1 \frac{1}{3}\ \ \Rightarrow\ \ m_1= -\frac{1}{6} \\\\y=2x-8\ \Rightarrow\ \ m_2= 2 \\\\m_1\cdot m_2= -\frac{1}{6} \cdot2=- \frac{1}{3} \neq -1$

7 0

3 years ago

Wish is 7(x-4) in simplest form?

butalik [34]

I'm pretty sure the attachment down there can answer ur question! :D

3 0

3 years ago

Read 2 more answers

The average yield of the standard variety off soybeans per acre on a farm in a region is 48.8 bushels per acre, with a standard

I am Lyosha [343]

Answer:

$z=\frac{53.4-48.8}{\frac{12}{\sqrt{36}}}=2.3$

$p_v =P(Z>2.3)=1-P(Z$

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significantly higher than 48.8 at 1% of signficance.

Step-by-step explanation:

1) Data given and notation

$\bar X=53.4$ represent the sample mean

$\sigma=12$ represent the population standard deviation assumed

$n=36$ sample size

$\mu_o =48.8$ represent the value that we want to test

$\alpha=0.01$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

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

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is higher than 48.8, the system of hypothesis would be:

Null hypothesis: $\mu \leq 48$

Alternative hypothesis: $\mu > 48$

Since we assume that know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (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".

3) Calculate the statistic

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

$z=\frac{53.4-48.8}{\frac{12}{\sqrt{36}}}=2.3$

4)P-value

Since is a right tailed test the p value would be:

$p_v =P(Z>2.3)=1-P(Z$

5) Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significantly higher than 48.8 at 1% of signficance.

3 0

3 years ago

The city of Orlando’s building a dog park in downtown Orlando dog park will be triangular shaped as shown on the coordinate plan

kiruha [24]

Answer:

Perimeter of the dog park = 15.4 yards

Step-by-step explanation:

Coordinates of the vertices of the given triangle are,

P(1, 2), Q(1, 6), R(-4, 2)

Since distance between the two points $(x_1, y_1)$ and $(x_2,y_2)$ is,