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3 years ago

The lines whose equations are 2x + 3y = 4 and y = mx + 6 will be perpendicular when

Mathematics

1 answer:

Natalka [10]3 years ago

5 0

Answer:

These lines will be perpendicular when m = ³⁄₂

Step-by-step explanation:

Two lines are perpendicular when the product of their gradients equal to -1. So:

m₁ * m₂ = -1

Let's get the first equation in gradient-intercept form (same as equation two).

2x + 3y = 4

3y = -2x + 4

y = -⅔x + ⁴⁄₃

We know the first gradient, and now we can find the gradient of the second line (which when multiplied will result in -1 as these lines are perpendicular). So:

-⅔ * m₂ = -1

m₂ = ³⁄₂

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Help w this pls it would save me<br> Pleasereeee

elena55 [62]

Answer:

0.6 over 0.6. Or just 1.

Step-by-step explanation:

5 0

3 years ago

Debra has ridden 57 miles of a bike course the course is 60 miles long what percentage of the course has Debra ridden so far

Reptile [31]

Hello!

To find our answer, we just divide.

57/60=0.95

We multiply by 100 to find our percent

100(0.95)-95

Therefore, our answer is 95%

I hope this helps!

4 0

4 years ago

Read 2 more answers

Solve the system of linear equations below.

Reika [66]

Answer:

x = 1 , y = 3 thus: A is your Anser

Step-by-step explanation:

Solve the following system:

{2 x + y = 5 | (equation 1)

x + y = 4 | (equation 2)

Subtract 1/2 × (equation 1) from equation 2:

{2 x + y = 5 | (equation 1)

0 x+y/2 = 3/2 | (equation 2)

Multiply equation 2 by 2:

{2 x + y = 5 | (equation 1)

0 x+y = 3 | (equation 2)

Subtract equation 2 from equation 1:

{2 x+0 y = 2 | (equation 1)

0 x+y = 3 | (equation 2)

Divide equation 1 by 2:

{x+0 y = 1 | (equation 1)

0 x+y = 3 | (equation 2)

Collect results:

Answer: {x = 1 , y = 3

8 0

3 years ago

Read 2 more answers

Suppose X, Y, and Z are random variables with the joint density function f(x, y, z) = Ce−(0.5x + 0.2y + 0.1z) if x ≥ 0, y ≥ 0, z

kompoz [17]

$f_{X,Y,Z}(x,y,z)=\begin{cases}Ce^{-(0.5x+0.2y+0.1z)}&\text{for }x\ge0,y\ge0,z\ge0\\0&\text{otherwise}\end{cases}$

is a proper joint density function if, over its support, $f$ is non-negative and the integral of $f$ is 1. The first condition is easily met as long as $C\ge0$ . To meet the second condition, we require

$\displaystyle\int_0^\infty\int_0^\infty\int_0^\infty f_{X,Y,Z}(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz=100C=1\implies \boxed{C=0.01}$

b. Find the marginal joint density of $X$ and $Y$ by integrating the joint density with respect to $z$ :

$f_{X,Y}(x,y)=\displaystyle\int_0^\infty f_{X,Y,Z}(x,y,z)\,\mathrm dz=0.01e^{-(0.5x+0.2y)}\int_0^\infty e^{-0.1z}\,\mathrm dz$

$\implies f_{X,Y}(x,y)=\begin{cases}0.1e^{-(0.5x+0.2y)}&\text{for }x\ge0,y\ge0\\0&\text{otherwise}\end{cases}$

Then

$\displaystyle P(X\le1.375,Y\le1.5)=\int_0^{1.5}\int_0^{1.375}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy$

$\approx\boxed{0.12886}$

c. This probability can be found by simply integrating the joint density:

$\displaystyle P(X\le1.375,Y\le1.5,Z\le1)=\int_0^1\int_0^{1.5}\int_0^{1.375}f_{X,Y,Z}(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz$

$\approx\boxed{0.012262}$

7 0

3 years ago

A professor using an open source introductory statistics book predicts that 10% of the students will purchase a hard copy of the

Likurg_2 [28]

Answer: The professor was not accurate with his hypothesis.

Null hypothesis: P1 = 12.5%, P2 = 42.5%, P3 = 45%

The alternate hypothesis: At least one proportion of the student will differ from the others.

Step-by-step explanation: To check if the professors hypothesis were inaccurate.

What percentage of student bought a hard copy of the book.

(25 ÷ 200) × 100 = 12.5%

What percentage of the student printed it from the web.

(85 ÷ 200) × 100 = 42.5%

What percentage of the students read it online.

(90 ÷ 200) × 100 = 45%

This means that the professor was not accurate with his hypothesis. Because the proportion of student in his hypothesis was not the same in the actual.

Therefore; the null hypothesis are

P1 = 12.5%, P2 = 42.5%, P3 = 45%

The alternative hypothesis will state that at least one of the proportion will be different from the others.

5 0

3 years ago