Ask question

Ask question

All categories

3 years ago

5

There are 10 girls and 6 boys in a class. The teacher randomly draws one student's name and then draws another. What is the prob

ability that the first name drawn will be a boy and the second will be a girl? (Please show work so I can understand)
60%
23.4%
16%
25%

2 answers:

Katen [24]3 years ago

8 0

Answer:

D 25%

Step-by-step explanation:

kumpel [21]3 years ago

6 0

Answer:

25%

Step-by-step explanation:

You might be interested in

A model car is 8 inches long and 5.2

Sphinxa [80]

Use proportions here - 8/5.2 = 96/width of actual car. using cross multiplication, 5.2*96 = 8x. this gives you 499.2 = 8x. divide both sides by eight to get x=62.4. the width of the actual car is 62.4 inches.

7 0

3 years ago

What is the solution to the system of equations graphed below

aliina [53]

The point where the lines intersect is the point that satisfies both equations.

It is (1, 5), selection C.

3 0

3 years ago

Read 2 more answers

Someone please I’m begging this is due by midnight and I can’t have it be late

Thepotemich [5.8K]

Answer:

(1) Coeficcient (2) Constant (3) 4 terms (4) In descending order, 2, 1, 3, 4 (5) -2 (6) 1 (7) 5 (8) 2 (9) -15, 5x, -7x^2, 2x^3 (10) -4x^2y, 2xy, 5x, 7

Step-by-step explanation:

Sorry if I got one wrong

8 0

3 years ago

Read 2 more answers

If cos() = − 2 3 and is in Quadrant III, find tan() cot() + csc(). Incorrect: Your answer is incorrect.

nydimaria [60]

Answer:

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}$

Step-by-step explanation:

Given

$\cos(\theta) = -\frac{2}{3}$

$\theta \to$ Quadrant III

Required

Determine $\tan(\theta) \cdot \cot(\theta) + \csc(\theta)$

We have:

$\cos(\theta) = -\frac{2}{3}$

We know that:

$\sin^2(\theta) + \cos^2(\theta) = 1$

This gives:

$\sin^2(\theta) + (-\frac{2}{3})^2 = 1$

$\sin^2(\theta) + (\frac{4}{9}) = 1$

Collect like terms

$\sin^2(\theta) = 1 - \frac{4}{9}$

Take LCM and solve

$\sin^2(\theta) = \frac{9 -4}{9}$

$\sin^2(\theta) = \frac{5}{9}$

Take the square roots of both sides

$\sin(\theta) = \±\frac{\sqrt 5}{3}$

Sin is negative in quadrant III. So:

$\sin(\theta) = -\frac{\sqrt 5}{3}$

Calculate $\csc(\theta)$

$\csc(\theta) = \frac{1}{\sin(\theta)}$

We have: $\sin(\theta) = -\frac{\sqrt 5}{3}$

So:

$\csc(\theta) = \frac{1}{-\frac{\sqrt 5}{3}}$

$\csc(\theta) = \frac{-3}{\sqrt 5}$

Rationalize

$\csc(\theta) = \frac{-3}{\sqrt 5}*\frac{\sqrt 5}{\sqrt 5}$

$\csc(\theta) = \frac{-3\sqrt 5}{5}$

So, we have:

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta)$

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \tan(\theta) \cdot \frac{1}{\tan(\theta)} + \csc(\theta)$

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = 1 + \csc(\theta)$

Substitute: $\csc(\theta) = \frac{-3\sqrt 5}{5}$

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = 1 -\frac{3\sqrt 5}{5}$

Take LCM

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}$

6 0

3 years ago

The jars of paint in the art room have different amounts of paint.the green paint jar is 4/8 full. The purple paint jar is 4/6 f

ruslelena [56]

The green paint is less full

8 0

3 years ago