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borishaifa [10]
3 years ago
14

six girls and four boys have entered the science fair. first, second, and third place awards are to be given out. what is the pr

obability that exactly one girl and two boys will recite awards? express your answer as a percent
Mathematics
2 answers:
Dafna1 [17]3 years ago
7 0
Given:
6 girls
4 boys

Probability of 1 girl being chosen: 1/6
Probability of 1 boy being chosen: 1/4
Probability of another boy being chosen: 1/3

1/6 = 0.17 x 100% = 17%
1/4 = 0.25 x 100% = 25%
1/3 = 0.33 x 100% = 33%

17% + 25% + 33% = 75%

The probability that exactly one girl and two boys will receive awards is 75%.
Ad libitum [116K]3 years ago
7 0
The probability that exactly one girl and two boys will receive awards is 75%. Have a good day!
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Use the expression to complete the statements. (0.5)10 (0.5)10 is the . (0.5) is the of (0.5)10. 10 is the of (0.5)10. A power i
RideAnS [48]

Question:

Use the expression to complete the statements.

(0.5)^{10\\

________ is the  power .

(0.5) is the  ________

A ______ is another way to write

Answer:

0.5 is the base

10 is the power

(0.5)^{10\\ = (\frac{1}{2})^{10} -- another way to write

Step-by-step explanation:

Given

(0.5)^{10\\

Required

Fill ion the gaps

(0.5)^{10\\ is pronounced as; 0.5 raise to the power of 10

From the above statement, we can deduce the following:

0.5 is the base

10 is the power

Lastly, another way to write (0.5)^{10\\

Express 0.5 as a fraction

(\frac{5}{10})^{10}

Simplify the fraction

(\frac{5/5}{10/5})^{10}

(\frac{1}{2})^{10}

Hence:

(0.5)^{10\\ = (\frac{1}{2})^{10}

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Step-by-step explanation:

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What is the solution of
kobusy [5.1K]

Answer:

Third option: x=0 and x=16

Step-by-step explanation:

\sqrt{2x+4}-\sqrt{x}=2

Isolating √(2x+4): Addind √x both sides of the equation:

\sqrt{2x+4}-\sqrt{x}+\sqrt{x}=2+\sqrt{x}\\ \sqrt{2x+4}=2+\sqrt{x}

Squaring both sides of the equation:

(\sqrt{2x+4})^{2}=(2+\sqrt{x})^{2}

Simplifying on the left side, and applying on the right side the formula:

(a+b)^{2}=a^{2}+2ab+b^{2}; a=2, b=\sqrt{x}

2x+4=(2)^{2}+2(2)(\sqrt{x})+(\sqrt{x})^{2}\\ 2x+4=4+4\sqrt{x}+x

Isolating the term with √x on the right side of the equation: Subtracting 4 and x from both sides of the equation:

2x+4-4-x=4+4\sqrt{x}+x-4-x\\ x=4\sqrt{x}

Squaring both sides of the equation:

(x)^{2}=(4\sqrt{x})^{2}\\ x^{2}=(4)^{2}(\sqrt{x})^{2}\\ x^{2}=16 x

This is a quadratic equation. Equaling to zero: Subtract 16x from both sides of the equation:

x^{2}-16x=16x-16x\\ x^{2}-16x=0

Factoring: Common factor x:

x (x-16)=0

Two solutions:

1) x=0

2) x-16=0

Solving for x: Adding 16 both sides of the equation:

x-16+16=0+16

x=16

Let's prove the solutions in the orignal equation:

1) x=0:

\sqrt{2x+4}-\sqrt{x}=2\\ \sqrt{2(0)+4}-\sqrt{0}=2\\ \sqrt{0+4}-0=2\\ \sqrt{4}=2\\ 2=2

x=0 is a solution


2) x=16

\sqrt{2x+4}-\sqrt{x}=2\\ \sqrt{2(16)+4}-\sqrt{16}=2\\ \sqrt{32+4}-4=2\\ \sqrt{36}-4=2\\ 6-4=2\\ 2=2

x=16 is a solution


Then the solutions are x=0 and x=16


5 0
3 years ago
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