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ASHA 777 [7]
3 years ago
5

A bag contains 6 red tiles, 8 blue tiles, and 6 green tiles. A tile is drawn, the color recorded, and then the tile placed back

in the bag before a tile is drawn again. If this procedure is done 30 times, how many red tiles would you expect to draw?
Mathematics
1 answer:
tester [92]3 years ago
3 0

Answer:

9 red tiles


Step-by-step explanation:


Total number of tiles: 6+8+6=20

Total number of red tiles: 6


Probability of getting red in one draw:  \frac{6}{20}=0.3

<em>This means that you would expect to get red 0.3  or 30% of the times.</em>

<em>Hence, when you draw tiles 30 times from the bag, you would expect red to be picked 30% of 30 times. This is:</em>

<em>0.3*30=9</em>

So, you would expect to draw a red tile from the bag 9 times.

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Whats the Approximate volume of the cylinder below in cubic centimeters?? Plz help
Sphinxa [80]

Answer:

Volume of a cylinder = 549.5 cm³

Step-by-step explanation:

Volume of a cylinder = πr²h

Where,

π = 3.14

Radius, r = 5 cm

Height, h = 7 cm

Volume of a cylinder = πr²h

= 3.14 * (5 cm)² * 7 cm

= 3.14 * 25 cm² * 7 cm

= 549.5 cm³

Volume of a cylinder = 549.5 cm³

3 0
2 years ago
8-x+4-2-8x= What does x equal
goldfiish [28.3K]

Answer:

-9x + 10

Step-by-step explanation:

3 0
3 years ago
I need help on this one
Tanzania [10]
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4 0
3 years ago
Which of the following functions have a vertical asymptote for values of Ø such that cos Ø = 1?Select all that apply.
igomit [66]

Answer:

y=\cot \theta and y=\csc \theta

Step-by-step explanation:

Note that if \cos \theta=1, then \sin \theta=0.

Functions y=\cos \theta,\ \ y=\sin \theta do not have vertical asymptotes at all.

Vertical asymptotes have functions y=\tan \theta,\ \ y=\cot \theta,\ \ y=\sec \theta,\ \ y=\csc \theta.

Functions y=\tan \theta and y=\sec \theta have the same vertical asymptotes (when \cos \theta =0).

Functions y=\cot \theta and y=\csc \theta have the same vertical asymptotes (when \sin \theta =0). See attached diagram

3 0
3 years ago
Pleas help me find the composition of goh and it’s domain
Ksivusya [100]

Solution:

The function is given below as

\begin{gathered} g(x)=\frac{x+4}{x-1} \\ h(x)=2x-1 \end{gathered}

To figure out

(goh)(x)

To do this , we will substitute x= 2x-1 in g(x)

\begin{gathered} g(x)=\frac{x+4}{x-1} \\ g(h)(x)=\frac{2x-1+4}{2x-1-1} \\ g(h)(x)=\frac{2x+3}{2x-2} \end{gathered}

Hence,

The composte function will be

(goh)(x)=\frac{2x+3}{2x-2}

Step 2:

To figure out the domain,

In mathematics, the domain of a function is the set of inputs accepted by the function.

Hence,

The domain of the function is

\begin{bmatrix}\mathrm{Solution:}\:&\:x1\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:1\right)\cup \left(1,\:\infty \:\right)\end{bmatrix}

5 0
1 year ago
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