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2 years ago

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4.

1 answer:

Roman55 [17]2 years ago

6 0

Part A:What percentage of the total students surveyed like both watching television and reading?

The total amount of student is 230
The amount of students that like to read and watch television is 80

As a fraction it is 80/230
Turn that to a percentage
It is:34.7826%

Part A=34.7826%

Part B:What is the probability that a student does not like watching television also does not like reading?

On the table it shows 20

So the probability is 20/230

Part B=20/230

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F(x) = 4/x-9<br> g(x)=4/x + 9<br> find f(g(x))

Alenkinab [10]

Hello,

$f(x)= \dfrac{4}{x-9} \\

g(x)= \dfrac{4}{x+9} \\

f(g(x))=f( \dfrac{4}{x+9} )\\

= \dfrac{4}{ \dfrac{4}{x+9}-9} \\

= \dfrac{4(x+9)}{4-9x-81} \\

=- \dfrac{4x+36}{77x+9x} 


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6 0

3 years ago

How do you do this and what ya the correct answer? Simple and concise explanation please!

ale4655 [162]

Answer:

2

Step-by-step explanation:

You are right as rain. You go to the x axis.

Find x = 3.5

f(3.5) is the y value of x = 3.5

f(3.5) = 2

It might help you a bit if you wrote it as a point (3.5,2)

4 0

3 years ago

A kitchen wall tile is shaped like a parallelogram. The base of the tile is two times the height of the tile . The height of the

SIZIF [17.4K]

Answer:

12.5 in²

Step-by-step explanation:

Given:

Height of tile = 2.5 in

Base of tile = 2*height of tile = 2*2.5 = 5 in

Length of other side of tile = 2.5 in

Required:

Area of the tile

SOLUTION:

Area of the tile = area of parallelogram

All we need to calculate the area of the tile, is the base and the height.

Area of parallelogram = base × height

Area of tile = 5 × 2.5 = 12.5 in²

5 0

3 years ago

Find the common ratio of the sequence 90, 60, 40, 80/3<br> What is the common ratio?

Trava [24]

Answer:

2/3

Step-by-step explanation:

<em>Common ratio = next term/previous term</em>

<u>Sequence:</u>

90, 60, 40, 80/3

<u>Common ratio:</u>

60/90 = 2/3

6 0

3 years ago

Find the exact values of a) sec of theta b)tan of theta if cos of theta= -4/5 and sin<0

Gre4nikov [31]

Answer:

Using trigonometric ratio:

$\sec \theta = \frac{1}{\cos \theta}$

$\tan \theta = \frac{\sin \theta}{\cos \theta}$

From the given statement:

$\cos \theta = -\frac{4}{5}$ and sin < 0

⇒ $\theta$ lies in the 3rd quadrant.

then;

$\sec \theta = \frac{1}{-\frac{4}{5}} = -\frac{5}{4}$

Using trigonometry identities:

$\sin \theta = \pm \sqrt{1-\cos^2 \theta}$

Substitute the given values we have;

$\sin \theta = \pm\sqrt{1-(\frac{-4}{5})^2 } =\pm\sqrt{1-\frac{16}{25}} =\pm\sqrt{\frac{25-16}{25}} =\pm \sqrt{\frac{9}{25} } = \pm\frac{3}{5}$

Since, sin < 0

⇒ $\sin \theta = -\frac{3}{5}$

now, find $\tan \theta$ :

$\tan \theta = \frac{\sin \theta}{\cos \theta}$

Substitute the given values we have;

$\tan \theta = \frac{-\frac{3}{5} }{-\frac{4}{5} } = \frac{3}{5}\times \frac{5}{4} = \frac{3}{4}$

Therefore, the exact value of:

(a)

$\sec \theta =-\frac{5}{4}$

(b)

$\tan \theta= \frac{3}{4}$

7 0

3 years ago