Find the equation of the line through (0,-2) and (4, 18).

What is 2/3rd a decimal rounded to the third decimal

dimulka [17.4K]

Answer:

0.667 or 0.666 (with the line over the last sixth)

Step-by-step explanation:

The original answer is 0.666666666666666 (continued on forever). Find the third decimal place and determine whether the next number follows the "5 or greater" rounding rule and round from there. The answer depends on the type of teacher.

5 0

3 years ago

You have 50$ to spend at the carnival. First, you must pay $9.20 for an entrance ticket. Each snack costs $4.25, including tax.

asambeis [7]

50 - 9.20 = 40.80 so 40.80/ 4.25 = 9.6 but you can’t buy .6 of a snack so it’s 9

5 0

3 years ago

Read 2 more answers

Simplify<br> X2-49<br> Over <br> X2+9x+14

Stels [109]

Answer:

(x-7)/(x+2)

Step-by-step explanation:

(x-7)(x+7)/(x+7)(x+2)

(x-7)/(x+2)

7 0

3 years ago

This is a 3 part 1 question each part on how to locate the plane and a small explanation report working out the recovery details

sesenic [268]

Third leg.

The crew flies at a speed of 560 mi/h in direction N-20°-E.

The wind has a speed of 35 mi/h and a direction S-10°-E.

We then can draw this as:

We have to add the two vectors to find the actual speed and direction.

We will start by adding the x-coordinate (W-E axis):

$\begin{gathered} x=560\cdot\sin (20\degree)+35\cdot\sin (10\degree) \\ x\approx560\cdot0.342+35\cdot0.174 \\ x\approx191.53+6.08 \\ x\approx197.61 \end{gathered}$

and the y-coordinate (S-N axis) is:

$\begin{gathered} y=560\cdot\cos (20\degree)-35\cdot\cos (10\degree) \\ y\approx560\cdot0.940-35\cdot0.985 \\ y\approx526.23-34.47 \\ y\approx491.76 \end{gathered}$

Then, the actual speed vector is v3=(197.61, 491.76).

The starting location for the third leg is R2=(216.66, 167.67) [taken from the previous answer].

Then, we have to calculate the displacement in 20 minutes using the actual speed vector.

We can calculate the movement in each of the axis. For the x-axis:

$\begin{gathered} R_{3x}=R_{2x}+v_{3x}\cdot t \\ R_{3x}=216.66+197.61\cdot\frac{1}{3} \\ R_{3x}=216.66+65.87 \\ R_{3x}=282.53 \end{gathered}$

NOTE: 20 minutes represents 1/3 of an hour.

We can do the same with the y-coordinate:

$\begin{gathered} R_{3y}=R_{2y}+v_{3y}\cdot t \\ R_{3y}=167.67+491.76\cdot\frac{1}{3} \\ R_{3y}=167.67+163.92 \\ R_{3y}=331.59 \end{gathered}$

The final position is R3 = (282.53, 331.59).

To find the distance from the origin and direction, we transform the cartesian coordinates of R3 into polar coordinates:

The distance can be calculated as if it was a right triangle:

$\begin{gathered} d^2=x^2+y^2_{} \\ d^2=282.53^2+331.59^2 \\ d^2=79823.20+109951.93 \\ d^2=189775.13 \\ d=\sqrt[]{189775.13} \\ d\approx435.63 \end{gathered}$

The angle, from E to N, can be calculated as:

$\begin{gathered} \tan (\alpha)=\frac{y}{x} \\ \tan (\alpha)=\frac{331.59}{282.53} \\ \tan (\alpha)\approx1.1736 \\ \alpha=\arctan (1.1736) \\ \alpha=49.56\degree \end{gathered}$

If we want to express it from N to E, we substract the angle from 90°:

$\beta=90\degree-\alpha=90-49.56=40.44\degree$

Answer: the final location can be represented with the vector (282.53, 331.59).

1) The distance from the origin is 435.63 miles and

2) the direction is N-40°-E.

7 0

1 year ago

A psychological experiment was conducted to investigate the length of time (time delay) between the administration of a stimulus

aleksklad [387]

Answer:

The P-value of this sample is 0 and is less than the significance level (0.05), so the effect is significant and the null hypothesis is rejected.

As there is significant evidence to reject $H_0: \mu= 1.6$ , we can say that there is significant evidence to claim that the mean time delay for the hypothetical population of all persons who may be subjected to the stimulus differs from 1.6 seconds.

The significance level for this test is 0.05.

Step-by-step explanation:

In this case, we want to prove if there is significant evidence that the mean differs from 1.6 seconds. That is the same as having evidence to reject the the hypothesis $H_0:\mu= 1.6$ (the null hypothesis always have the equal sign).

We have to test the null hypothesis

$H_0:\mu= 1.6$

The significance level for this test is 0.05

Calculation of the t-statistic:

$t=\frac{M-\mu}{s/\sqrt{n}} =\frac{2.2-1.6}{0.57/\sqrt{36}} =\frac{0.6}{0.063}= 9.47$

If we look up in a t-table for t=9.47 and df=(36-1)=35, we get this value appears with a probability of zero. A large number like that is very unlikely to happen.

The P-value of this sample is 0 and is less than the significance level (0.05), so the effect is significant and the null hypothesis is rejected.

As there is significant evidence to reject $H_0: \mu= 1.6$ , we can say that there is significant evidence to claim that the mean time delay for the hypothetical population of all persons who may be subjected to the stimulus differs from 1.6 seconds.

6 0

3 years ago