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3 years ago

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In a certain lottery, 3 balls are drawn (at random) from 10 white balls numbered from 1 to 10, and one special ball is drawn (at

random) from ten red balls numbered from 11 to 20. When you buy a ticket, you choose three numbers from 1 to 10, and one number from 11 to 20.
If the numbers on your ticket match at least two of the white balls or match the red SuperBall, then you win a super prize. What is the probability that you win a super prize?

If you answer this problem correctly, I will mark your answer as brainliest and you will get 50 points.

1 answer:

prohojiy [21]3 years ago

4 0

Answer:

What are the answer options... I have an answer but I want to check if it is correct... thanks

Step-by-step explanation:

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Solve 12^x^2+5x-4 = 12^2x+6

faust18 [17]

The solutions for ‘x’ are 2 and -5

<u>Step-by-step explanation:</u>

Given equation:

$12^{x^{2}+5 x-4}=12^{2 x+6}$

Since the base on both sides as ‘12’ are the same, we can write it as

$x^{2}+5 x-4=2 x+6$

$x^{2}+5 x-2 x-4-6=0$

$x^{2}+3 x-10=0$

Often, the value of x is easiest to solve by $a x^{2}+b x+c=0$ by factoring a square factor, setting each factor to zero, and then isolating each factor. Whereas sometimes the equation is too awkward or doesn't matter at all, or you just don't feel like factoring.

<u>The Quadratic Formula:</u> For $a x^{2}+b x+c=0$ , the values of x which are the solutions of the equation are given by:

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

Where, a = 1, b = 3 and c = -10

$x=\frac{-3 \pm \sqrt{(-3)^{2}-4(1)(-10)}}{2(1)}$

$x=\frac{-3 \pm \sqrt{9+40}}{2}$

$x=\frac{-3 \pm \sqrt{49}}{2}=\frac{-3 \pm 7}{2}$

So, the solutions for ‘x’ are

$x=\frac{-3+7}{2}=\frac{4}{2}=2$

$x=\frac{-3-7}{2}=\frac{-10}{2}=-5$

The solutions for ‘x’ are 2 and -5

6 0

3 years ago

Read 2 more answers

What is the vertices of hexagon<br>

nata0808 [166]

Number of vertices of hexagon is 6

5 0

3 years ago

Read 2 more answers

50 POINTS!!! In rectangle ABCD, AB = 6 cm, BC = 8 cm, and DE = DF. The area of triangle DEF is one-fourth the area of rectangle

aalyn [17]

Answer:

$EF=4\sqrt{3}$

Step-by-step explanation:

In rectangle ABCD, AB = 6, BC = 8, and DE = DF.

ΔDEF is one-fourth the area of rectangle ABCD.

We want to determine the length of EF.

First, we can find the area of the rectangle. Since the length AB and width BC measures 6 by 8, the area of the rectangle is:

$A_{\text{rect}}=8(6)=48\text{ cm}^2$

The area of the triangle is 1/4 of this. Therefore:

$\displaystyle A_{\text{tri}}=\frac{1}{4}(48)=12\text{ cm}^2$

The area of a triangle is half of its base times its height. The base and height of the triangle is DE and DF. Therefore:

$\displaystyle 12=\frac{1}{2}(DE)(DF)$

Since DE = DF:

$24=DF^2$

Thus:

$DF=\sqrt{24}=\sqrt{4\cdot 6}=2\sqrt{6}=DE$

Since ABCD is a rectangle, ∠D is a right angle. Then by the Pythagorean Theorem:

$(DE)^2+(DF)^2=(EF)^2$

Therefore:

$(2\sqrt6)^2+(2\sqrt6)^2=EF^2$

Square:

$24+24=EF^2$

Add:

$EF^2=48$

And finally, we can take the square root of both sides:

$EF=\sqrt{48}=\sqrt{16\cdot 3}=4\sqrt{3}$

6 0

3 years ago

Read 2 more answers

The equation of line a is y=12x−1, and it passes through point (6,2).

Triss [41]

Answer:

A). slope = -1/12 y-intercept = 3/2

Step-by-step explanation:

y = -1/12x + b

2 = -1/12(-6) + b

2 = 1/2 + b

3/2 = b

y = -1/12x + 3/2

3 0

3 years ago

Please help........................

VashaNatasha [74]

Answer:

<h2>m∠ABD = 20°</h2>

Step-by-step explanation:

If m∠ABC = 40° and BD is the bisector of ∠ABC, then

(1) m∠ABD = m∠DBC

(2) m∠ABC = m∠ABD + m∠DBC

From (1) and (2) we have:

m∠ABC = 2m∠ABD

Therefore

2m∠ABD = 40° <em>divide both sides by 2</em>

m∠ABD = 20°

3 0

3 years ago