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Jet001 [13]
2 years ago
6

After six rolls of a standard die, the experimental probability of rolling a 3 is 26. What do you expect will happen to the expe

rimental probability if the die is rolled 90 more times? Explain.
Mathematics
1 answer:
Anettt [7]2 years ago
6 0

Answer:

The experimental probability should get closer to the theoretical probability of 1/6 with more trials.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

Experimental probability:

The number of desired outcomes is taken from the results of an experiment.

Theoretical probability:

Found before the experiment happens.

For a large number of trials, the experimental probability will be closer to the theoretical probability.

In this question:

A standard die has 6 sides, one which is 3. So the theoretical probability of rolling a 3 is 1/6.

After six rolls of a standard die, the experimental probability of rolling a 3 is 2/6.

The experimental probability, after six rolls, is 2/6 = 1/3.

What do you expect will happen to the experimental probability if the die is rolled 90 more times?

As the number of trials increase, the experimental probability is expected to get closer to the theoretical probability, which in this case is 1/6.

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madam [21]

Answer:

18b - 15

Step-by-step explanation:

<u>Step 1:  Distribute </u>

-3(-6b + 5)

(-3 * -6b) + (-3 * 5)

18b - 15

Answer:   18b - 15

<u>If needed solve for b </u>

18b - 15 + 15 = 0 + 15

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3 years ago
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Step-by-step explanation:

31.32

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Please help me ----------------
Brut [27]

Answer:

C = 43.98 in

Step-by-step explanation:

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7 0
3 years ago
The perimeter of a rectangle is 30.8 km and it’s diagonal length is 11 km. Find it’s length and width
blsea [12.9K]

Answer:

Length of the rectangle is 15.0325 km and width is 0.3765 km.

Explanation:

Given:

Perimeter of a rectangle = 30.8 km

Length of diagonal of rectangle = 11 km

To find:

The length and width of rectangle=?

Solution:

Lets assume length of the rectangle = x km

And assume width of the rectangle = y km

Lets first create equation using given  perimeter

perimeter of rectangle = 2 ( length +  width )

=> 30.8 km = 2 ( x + y )  

=>x + y = \frac{30.8}{2}

=> y = 15.4 – x             ------(1)

As diagonal and two sides of rectangle forms right angle triangle whose hypoteneus is diagonal ,  

=> length^2 + width^2 = diagonal^2

=> x^2 + y^2 = 11^2

=> x^2 + y^2 = 121

On substituting value of y from (1) in above equation we get

=> x^2 + (15.4-x)^2 = 121

=>x^2 + (15.4)^2 + x^2 – 2 x 15.4 \times x   = 121

=> 2x^2-30.8x + 237.16 -121  = 0

=> 2x^2-30.8x + 116.16 = 0

Solving above quadratic equation using quadratic formula

General form of quadratic equation is  

ax^2 +bx +c = 0

And quadratic formula for getting roots of quadratic equation is  

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

As equation is 2x^2-30.8x + 116.16 = 0, in our case

a = 2 ,  b = -30.8 and c = 116.16

Calculating roots of the equation we get

x=\frac{ -(-30.8)\pm\sqrt{(-30.8)^2-4(2)( 11)} } {(2\times2)}

x=\frac{30.8\pm\sqrt{(948.64-88)}}{4}

x=\frac{30.8\pm\sqrt{860.64}}{4}

x=\frac{30.8\pm\sqrt{860.64}}{4}

x=\frac{(30.8\pm29.33)}4

x=\frac{(30.8+29.33)}{4}

x=\frac{(30.8-29.33)}{4}

=> x = 15.0325 or x = 0.3675

As generally length is longer one ,  

So x = 1.0325

From equation (1) y = 15.4 – x = 0.3765

Hence length of the rectangle is 15.0325 km and width is 0.3765 km.

6 0
3 years ago
I need answers for number 20 I need This done by Tuesday
nikdorinn [45]

One ticket costs $24.50 - $19 more for evening

Six tickets cost 6 (  $24.50 - $19 ) more

Answer: 6 (  $24.50 - $19 )

Problem 18 is D by the way

6 0
3 years ago
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