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3 years ago

How do you write equations for three lines that contain the point (0, 2)?

1 answer:

7 0

First we see points like (0,2) in place x is 0 and y is 2 so we write equation is 0 +2=2 so, x+y= 2
2×0+2= 2 so, 2x + y = 2
0 + 2×2=4 so, x+ 2y =4

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Last Wednesday, a random sample of 24 students were surveyed to find how long it takes to walk from the Fretwell Building to the

Ray Of Light [21]

Answer:

Step-by-step explanation:

Solution:-

- We are to investigate the confidence interval of 95% for the population mean of walking times from Fretwell Building to the college of education building.

- The survey team took a sample of size n = 24 students and obtained the following results:

Sample mean ( x^ ) = 12.3 mins

Sample standard deviation ( s ) = 3.2 mins

- The sample taken was random and independent. We can assume normality of the sample.

- First we compute the critical value for the statistics.

- The z-distribution is a function of two inputs as follows:

Significance Level ( α / 2 ) = ( 1 - CI ) / 2 = 0.05/2 = 0.025

Compute: z-critical = z_0.025 = +/- 1.96

- The confidence interval for the population mean ( u ) of walking times is given below:

[ x^ - z-critical*s / √n , x^ + z-critical*s / √n ]

Answer: [ 12.3 - 1.96*3.2 / √24 , 12.3 + 1.96*3.2 / √24 ]

3 0

3 years ago

Now there is a square city of unknown size with a gate at the center of each side. There is a tree 20 b from the north gate. Tha

Brums [2.3K]

Answer:

The length of each side of the city is 250b

Step-by-step explanation:

Given

$a = 20$ --- tree distance from north gate

$b =14$ --- movement from south gate

$c = 1775$ --- movement in west direction from (b)

See attachment for illustration

Required

Find x

To do this, we have:

$\triangle ADE \sim \triangle ACB$ --- similar triangles

So, we have the following equivalent ratios

$AE:DE = AB:CB$

Where:

$AE = 20\\ DE = x/2 \\ AB = 20 + x + 14 \\ CB = 1775$

Substitute these in the above equation

$20:x/2 = 20 + x + 14: 1775$

$20:x/2 = x + 34: 1775$

Express as fraction

$\frac{20}{x/2} = \frac{x + 34}{1775}$

$\frac{40}{x} = \frac{x + 34}{1775}$

Cross multiply

$x *(x + 34) = 1775 * 40$

Open bracket

$x^2 + 34x = 71000$

Rewrite as:

$x^2 + 34x - 71000 = 0$

Expand

$x^2 + 284x -250x - 71000 = 0$

Factorize

$x(x + 284) -250(x + 284)= 0$

Factor out x + 284

$(x - 250)(x + 284)= 0$

Split

$x - 250 = 0 \ or\ x + 284= 0$

Solve for x

$x = 250 \ or\ x =- 284$

x can't be negative;

So:

$x = 250$

6 0

3 years ago

Let’s now add x to both sides of the equation and consider the new equation x squared+5+x=11

irina1246 [14]

Answer:x=3

Step-by-step explanation:x squared is one x.you then add the X's to get 2x.then you subtract the five on the other side to get 2x=6.after you get that you divide both sides by 2 to get x=3

3 0

3 years ago

What is the distance between the points (1, –6) and (–5, 2)? Round your answer to the nearest tenth. A. 10.0 units B. 9.0 units

Mama L [17]

B .....................................................

3 0

3 years ago

if the sum of the first 60 positive integers is s, what is the sum of the first 120 integers in terms of s? a. 2s 3600 b. s^2 36

kvasek [131]

For the first 60 positive integers, a = 1, n = 60, l = 60.
Sn = n/2(a + l)
s = 60/2(1 + 60) = 30(61)

For the next 60 positive integer, a = 61, n = 60, l = 120
Sum = 60/2(61 + 120) = 30(61 + 120) = 30(61) + 30(120) = s + 3600

Sum of first 120 positive integers = s + s + 3600 = 2s + 3600

6 0

3 years ago