All categories

3 years ago

A line segment of length k is divided into 3 equal parts what is the distance between midpoints of the first and third segments

Mathematics

1 answer:

anastassius [24]3 years ago

4 0

Line segment of length k is divided into 3 equal parts.

so first segment is 0-k/3 and third segment is 2/3k-k

so mid-pt of 1st = k/6 and 3rd = 5/6k

so the distance in between = 5/6k-k/6 = 4/6k = 2/3k

You might be interested in

Functions 1 and 2 are shown below: Function 1: f(x) = −3x2 + 2 A graph of a parabola with x intercepts of negative 0.5, 0 and 2,

ANEK [815]

2. i belive the answer should be

6 0

3 years ago

Read 2 more answers

I need help please I don’t understand stand

vovikov84 [41]

Answer:

Step-by-step explanation:

all you gotta do is move the figure up 3 squares, and to the right 3 squares, the way I do this is one point at a time

8 0

3 years ago

Read 2 more answers

How many solutions does 5 - 2x = 3 + x-4- 3x have?

aev [14]

Answer:

Step-by-step explanation:

just took the test

6 0

3 years ago

Read 2 more answers

Asap need some help on this question<br> fast answer please thanks :)

stira [4]

Answer:

£ 550

Step-by-step explanation:

C = p(t + 30)

C = 2(245 + 30)

C = 2 (275) [using order of operations]

C = 550

Hope this helps!

4 0

2 years ago

Read 2 more answers

You must estimate the mean temperature (in degrees Fahrenheit) with the following sample temperatures: 44 32.8 59.2 31.4 12.7 68

Licemer1 [7]

Answer:

$51.278 -2.896 \frac{22.979}{\sqrt{9}}= 29.096$

$51.278 +2.896 \frac{22.979}{\sqrt{9}}= 73.460$

And the interval would be:

$(29.10 \leq \mu \leq 73.46)$

Step-by-step explanation:

For this problem we have the following dataset given:

44 32.8 59.2 31.4 12.7 68.5 84.7 72.5 55.7

We can find the mean and sample deviation with the following formulas:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And replacing we got:

$\bar X= 51.278$

$s= 22.979$

The confidence interval for the mean is given by:

$\bar X \pm t_{\alpha/2} \frac{s}{\sqrt{n}}$

The degrees of freedom are:

$df=n-1= 9-1=8$

The confidence would be 0.98 and the significance $\alpha=0.02$ then the critical value would be:

$t_{\alpha/2}= 2.896$

Ad replacing we got:

$51.278 -2.896 \frac{22.979}{\sqrt{9}}= 29.096$

$51.278 +2.896 \frac{22.979}{\sqrt{9}}= 73.460$

And the interval would be:

$(29.10 \leq \mu \leq 73.46)$

7 0

3 years ago