The length and the width of a rectangle are in the ratio of 5 to 2, and the area is 1,000 sq. ft. Find the dimensions.

PLS HELP ME WITH 12 and the bottom one ASAP!! (SHOW WORK FOR BOTH!!) + LOTS OF POINTS!!!

alexandr1967 [171]

Well, since its isoceles triangle so two equal sides. its either 3 or 7. There is a theorem that states two sides added must be greater than the third. so if the third side is 3. it would be 3+3=6, less than 7, incorrect. therefore the correct side is 7. so the sides are 3, 7, 7. you can add them to confirm the theorem.

3 0

3 years ago

Asystem of equations has 1 solution. If 4x - y = 5 is one of the equations, which could be the other equation?

NeX [460]

The answer for the problem would be A. Y = 4x + 5

8 0

3 years ago

How do you do these problems?

son4ous [18]

3.) 3 faces. 4 edges. 3 V
4.) 5 faces. 5 edges. 3 V
5.) 2 faces. 1 edge. 1 V
6.) 3 faces. 0 edges. 1 V
7.) 1 face. 0 edges. 2 V's
I'm iffy on the V's.

8 0

3 years ago

A discuss moves from P1 (4,8) to P2 (15,17). What is the lincar displacement in the horizontal and vertical directions? What is

Yakvenalex [24]

Answer:

The horizontal displacement is 11 units, the vertical displacement is 9 units, and the projection angle is 39.3 degrees.

Step-by-step explanation:

We can start using the definition of displacement in one dimension between any 2 points which is the difference between them, so we have

$\Delta s = s_2-s_1$

And apply it to get the horizontal and vertical displacements.

Once we have found them, we can use trigonometric functions to find the projection angle with respect the horizontal.

Linear displacements.

Using the definition of displacement, we can write the horizontal displacement as

$\Delta x = x_2-x_1$

So we can use the given points $P1:(x_1,y_2) \text{ and } P_2: (x_2,y_2)$ on the displacement formula

$\Delta x = 15-4\\\Delta x = 11$

In the same manner we can look at the y components of those points to find the vertical displacement

$\Delta y = 17-8\\\Delta y =9$

Thus the horizontal displacement is 11 units and the vertical displacement is 9 units.

Projection angle.

The projection angle with respect the horizontal is the angle that is made between the line that connects the points P1 and P2 and the horizontal, so we can use the linear displacements previously found to write

$\tan(\theta) = \cfrac{\Delta y}{\Delta x}$

Solving for the angle we get

$\theta = \tan^{-1}\left(\cfrac{\Delta y}{\Delta x}\right)$

Replacing values

$\theta = \tan^{-1}\left(\cfrac{9}{11}\right)$

Which give us

$\theta = 39.3^\circ$

So the projection angle is 39.3 degrees.

7 0

3 years ago

A business school has a goal that the average number of years of work experience of its MBA applicants is more than three years.

gizmo_the_mogwai [7]

Answer:

$z=\frac{3.1-3}{\frac{2.449}{\sqrt{47}}}=0.280$

$p_v =P(z>0.280)=0.390$

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 height of men actually its NOT significant higher than 0.3 at 1% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=3.1$ represent the sample mean

$\sigma=\sqrt{6}$ represent the sample standard deviation for the sample

$n=47$ sample size

$\mu_o =3$ 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 higher than 3, the system of hypothesis would be:

Null hypothesis: $\mu \leq 3$

Alternative hypothesis: $\mu > 3$

If we analyze the size for the sample is > 30 and we know the population deviation so 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".

Calculate the statistic

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

$z=\frac{3.1-3}{\frac{2.449}{\sqrt{47}}}=0.280$

P-value

Since is a one side test the p value would be:

$p_v =P(z>0.280)=0.390$

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 height of men actually its NOT significant higher than 0.3 at 1% of signficance.

3 0

3 years ago