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3 years ago

If (10,3) and (6,31) are two anchor points on a trend like, then find the equation of the line

Mathematics

1 answer:

Misha Larkins [42]3 years ago

8 0

Answer:

let's find the gradient first. (m)

Step-by-step explanation:

(10,3) (6,31)

y²-y¹/x²-x¹

31-3/6-3

28/3

(m) =28/3

y-y¹=m(x-x¹)

let's take y¹=3, x¹=10 and m= 28/3

y-3=28/3(x -10)

y-3=(28/3)x -128/3

taking LCM=3

3y-9=28x-128

3y=28x -128+9

3y=28x-119

Soo your equation is:

3y=28x-119

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Researchers are studying the distribution of subscribers to a certain streaming service in different populations. From a random

Katen [24]

Answer:

$CI = (0.17 - 0.27)\± 1.65\sqrt{\frac{(0.17)*(0.83) + (0.27)*(0.73)}{200}}$

Step-by-step explanation:

Given

$n = 200$

$x_1 = 34$ -- City C

$x_2 = 54$ --- City K

Required

Determine the 90% confidence interval

This is calculated using:

$CI = \bar x \± z\frac{\sigma}{\sqrt n}$

Calculating $\bar x$

$\bar x = \bar x_1 - \bar x_2$

$\bar x = \frac{x_1}{n} - \frac{x_2}{n}$

$\bar x = \frac{34}{200} - \frac{54}{200}$

$\bar x = 0.17 - 0.27$

For a 90% confidence level, the z-score is 1.65. So:

$z = 1.65$

Calculating the standard deviation $\sigma$

$\sigma = \sqrt{(\bar x_1)*(1 - \bar x_1) + (\bar x_2)*(1 - \bar x_2) }$

So:

$\sigma = \sqrt{(0.17)*(1 - 0.17) + (0.27)*(1 - 0.27) }$

$\sigma = \sqrt{(0.17)*(0.83) + (0.27)*(0.73)}$

So:

$CI = (0.17 - 0.27)\± 1.65\frac{\sqrt{(0.17)*(0.83) + (0.27)*(0.73)}}{\sqrt {200}}$

$CI = (0.17 - 0.27)\± 1.65\sqrt{\frac{(0.17)*(0.83) + (0.27)*(0.73)}{200}}$

7 0

3 years ago

In JKL, k=4.1 cm,j=3.8 cm and angle J=Q3^ Find all possible values of angle K , to the nearest inch of a degree .

spayn [35]

Answer:

74.0°

Step-by-step explanation:

In triangle JKL, k = 4.1 cm, j = 3.8 cm and ∠J=63°. Find all possible values of angle K, to the nearest 10th of a degree

Solution:

A triangle is a polygon with three sides and three angles. Types of triangles are right angled triangle, scalene triangle, equilateral triangle and isosceles triangle.

Given a triangle with angles A, B, C and the corresponding sides opposite to the angles as a, b, c. Sine rule states that for the triangle, the following holds:

$\frac{a}{sin(A)}=\frac{b}{sin(B)}=\frac{c}{sin(C)}$

In triangle JKL, k=4.1 cm, j=3.8 cm and angle J=63°.

Using sine rule, we can find ∠K:

$\frac{k}{sin(K)}=\frac{j}{sin(J)} \\\\\frac{4.1}{sin(K)}=\frac{3.8}{sin(63)} \\\\sin(K)=\frac{4.1*sin(63)}{3.8}\\\\sin(K)=0.9613\\\\K=sin^{-1} (0.9613)\\\\K=74.0^o \\$