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3 years ago

10 points.

Mathematics

1 answer:

rusak2 [61]3 years ago

5 0

Answer:

<u>The difference between the experimental probability and the theoretical probability is 0.08 (rounding the answer to the nearest hundredth) or 8%.</u>

Correct statement and question:

Regina has a bag of 6 orange marbles and 6 black marbles. She picks a marble at random and then puts it back in the bag. She does this 24 times. The results can be found in the table.

Outcome Tally

Orange 10

Black 14

Figure out the percent error of pulling a black marble in Regina’s experiment. Show your work and round the answer to the nearest hundredth.

Source:

Previous question that can be found at brainly

Step-by-step explanation:

1. Let's review the information provided to us to answer the question correctly:

Number of orange marbles = 6

Number of black marbles = 6

Number of times of the experiment = 24

Number of times the outcome was an orange marble = 10

Number of times the outcome was a black marble = 14

2. Figure out the percent error of pulling a black marble in Regina’s experiment. Show your work and round the answer to the nearest hundredth.

The theoretical probability of pulling a black marble is 12/24 or 0.5, given that the number of orange and black marbles are equal and actually, the experimental probability is 14/24 or 0.5833.

We can't describe this difference as an "error". What happened here is that there is a difference between the experimental probability and the theoretical probability.

0.5833 - 0.5 = 0.0833 = 0.08 (rounding the answer to the nearest hundredth)

<u>The difference between the experimental probability and the theoretical probability is 0.08 (rounding the answer to the nearest hundredth) or 8%.</u>

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Answer:

$(x+\frac{b}{2a})^2+(\frac{4ac}{4a^2}-\frac{b^2}{4a^2})=0$

$(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}$

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

Step-by-step explanation:

$x^2+\frac{b}{a}x+\frac{c}{a}=0$

They wanted to complete the square so they took the thing in front of x and divided by 2 then squared. Whatever you add in, you must take out.

$x^2+\frac{b}{a}x+(\frac{b}{2a})^2+\frac{c}{a}-(\frac{b}{2a})^2=0$

Now we are read to write that one part (the first three terms together) as a square:

$(x+\frac{b}{2a})^2+\frac{c}{a}-(\frac{b}{2a})^2=0$

I don't see this but what happens if we find a common denominator for those 2 terms after the square. (b/2a)^2=b^2/4a^2 so we need to multiply that one fraction by 4a/4a.

$(x+\frac{b}{2a})^2+\frac{4ac}{4a^2}-\frac{b^2}{4a^2}=0$

They put it in ( )

$(x+\frac{b}{2a})^2+(\frac{4ac}{4a^2}-\frac{b^2}{4a^2})=0$

I'm going to go ahead and combine those fractions now:

$(x+\frac{b}{2a})^2+(\frac{-b^2+4ac}{4a^2})=0$

I'm going to factor out a -1 in the second term ( the one in the second ( ) ):

$(x+\frac{b}{2a})^2-(\frac{b^2-4ac}{4a^2})=0$

Now I'm going to add (b^2-4ac)/(4a^2) on both sides:

$(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}$

I'm going to square root both sides to rid of the square on the x+b/(2a) part:

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

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

Now subtract b/(2a) on both sides:

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

Combine the fractions (they have the same denominator):

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

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3 years ago

*20 POINTS*

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Answer:

If you want to use the Rational Zeros Theorem, as instructed, you need to use synthetic division to find zeros until you get a quadratic remainder.

P: ±1, ±2, ±3, ±6 (all prime factors of constant term)

Q: ±1, ±7 (all prime factors of the leading coefficient)

P/Q: ±1, ±2, ±3, ±6, ±1/7, ±2/7, ±3/7, ±6/7 (all possible values of P/Q)

Now, start testing your values of P/Q in your polynomial:

f(x)=7x4-9x3-41x2+13x+6

You can tell f(1) and f(-1) are not zeros since they're not = 0Now try f(2) and f(-2):

f(2)=7(16)-9(8)-41(4)+13(2)+6

112-72-164+26+6 ≠ 0

f(-2)=7(16)-9(-8)-41(4)+13(-2)+6

112+72-164-26+6 = 0 OK!! There is a zero at x=-2

This means (x+2) is a factor of the polynomial.

Now, do synthetic division to find the polynomial that results from

(7x4-9x3-41x2+13x+6)÷(x+2):

-2⊥ 7 -9 -41 13 6

-14 46 -10 -6

7 -23 5 3 0 The remainder is 0, as expected

The quotient is a polynomial of degree 3:

7x3-23x2+5x+3

Now, continue testing the P/Q values with this new polynomial. Try f(3):

f(3)=7(27)-23(9)+5(3)+3

189-207+15+3 = 0 OK!! we found another zero at x=3

Now, another synthetic division:

3⊥ 7 -23 5 3

21 -6 -3_

7 -2 -1 0

The quotient is a quadratic polynomial:

7x2-2x-1 This is not factorable, you need to apply the quadratic formula to find the 3rd and 4th zeros:

x= (1±2√2)÷7

The polynomial has 4 zeros at x=-2, 3, (1±2√2)÷7

Step-by-step explanation:

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