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Inessa05 [86]
3 years ago
5

There are slips of paper that are numbered from 1 to 15 in a hat. What is the theoretical probability that a slip of paper that

is chosen will be an even number?
1/15
7/15
1/2
8/15
Mathematics
2 answers:
blsea [12.9K]3 years ago
8 0
Okay! Let's write out all the numbers: 
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15

Now, let's bold (or you can circle them... or do really anything to them as long as they are standing out to you) the even numbers (remember all even numbers are divisible by 2)

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15

Now let's list all the ones we highlighted: 
2
4
6
8
10
12
14

Count up how many numbers that is. 
You should get 7. 

So, our answer is 7 out of 15 numbers. 
Or as a fraction: 

7/15 is our answer.

Note: A probability is the number of chances something has to happen out of the total number of chances. So in this case the number of even numbers out of the total number of numbers. You can apply this to any probability problem. Hope this helps!

DochEvi [55]3 years ago
6 0
There are 7 even numbers between 1 and 15 so the answer will be 7/15

Hope this helps :)
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Eva8 [605]
9/15 becomes 90/15 but now u have a point on ur answer so it would be .6 as 15 times 6 equals 90
4 0
3 years ago
Read 2 more answers
A box with a hinged lid is to be made out of a rectangular piece of cardboard that measures 3 centimeters by 5 centimeters. Six
kherson [118]

Answer:

x = 0.53 cm

Maximum volume = 1.75 cm³

Step-by-step explanation:

Refer to the attached diagram:

The volume of the box is given by

V = Length \times Width \times Height \\\\

Let x denote the length of the sides of the square as shown in the diagram.

The width of the shaded region is given by

Width = 3 - 2x \\\\

The length of the shaded region is given by

Length = \frac{1}{2} (5 - 3x) \\\\

So, the volume of the box becomes,

V =  \frac{1}{2} (5 - 3x) \times (3 - 2x) \times x \\\\V =  \frac{1}{2} (5 - 3x) \times (3x - 2x^2) \\\\V =  \frac{1}{2} (15x -10x^2 -9 x^2 + 6 x^3) \\\\V =  \frac{1}{2} (6x^3 -19x^2 + 15x) \\\\

In order to maximize the volume enclosed by the box, take the derivative of volume and set it to zero.

\frac{dV}{dx} = 0 \\\\\frac{dV}{dx} = \frac{d}{dx} ( \frac{1}{2} (6x^3 -19x^2 + 15x)) \\\\\frac{dV}{dx} = \frac{1}{2} (18x^2 -38x + 15) \\\\\frac{dV}{dx} = \frac{1}{2} (18x^2 -38x + 15) \\\\0 = \frac{1}{2} (18x^2 -38x + 15) \\\\18x^2 -38x + 15 = 0 \\\\

We are left with a quadratic equation.

We may solve the quadratic equation using quadratic formula.

The quadratic formula is given by

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Where

a = 18 \\\\b = -38 \\\\c = 15 \\\\

x=\frac{-(-38)\pm\sqrt{(-38)^2-4(18)(15)}}{2(18)} \\\\x=\frac{38\pm\sqrt{(1444- 1080}}{36} \\\\x=\frac{38\pm\sqrt{(364}}{36} \\\\x=\frac{38\pm 19.078}{36} \\\\x=\frac{38 +  19.078}{36} \: or \: x=\frac{38 - 19.078}{36}\\\\x= 1.59 \: or \: x = 0.53 \\\\

Volume of the box at x= 1.59:

V =  \frac{1}{2} (5 – 3(1.59)) \times (3 - 2(1.59)) \times (1.59) \\\\V = -0.03 \: cm^3 \\\\

Volume of the box at x= 0.53:

V =  \frac{1}{2} (5 – 3(0.53)) \times (3 - 2(0.53)) \times (0.53) \\\\V = 1.75 \: cm^3

The volume of the box is maximized when x = 0.53 cm

Therefore,

x = 0.53 cm

Maximum volume = 1.75 cm³

7 0
3 years ago
Which explains the error Krista made?
MariettaO [177]
A is the right option! Because 0.40 should be in second equation with the x to represent the number of stamps multiply by 40 cents.
4 0
3 years ago
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Harman [31]

\to \sf(3x^3-4-3x) - (-8+2x^3)

\to \sf(3x^3-4-3x)  + 8 - 2x^3

\to \sf3x^3-4-3x  + 8 - 2x^3

\to \sf3x^3 - 2x^3-3x  +4

\to \sf x^3-3x  +4

3 0
3 years ago
Read 2 more answers
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Nonamiya [84]

Answer:

i will give it to you

Step-by-step explanation:

that's it for you dear

7 0
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