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Inga [223]
4 years ago
12

a group of students is donating blood during a blood drive. a student has 9/20 probability of having type 'o' blood and a 2/5 pr

obability of having type 'A'.what is the probability that a student has type 'o' or type 'a' blood and why?
Mathematics
2 answers:
Anna [14]4 years ago
7 0
Hello there.
<span>
A group of students is donating blood during a blood drive. a student has 9/20 probability of having type 'o' blood and a 2/5 probability of having type 'A'.what is the probability that a student has type 'o' or type 'a' blood and why?
</span>
The student can have only one blood type, so the actual events are mutually exclusive.
guajiro [1.7K]4 years ago
6 0
For the answer to the question above,
the answer is "<span>The student can have only one blood type, so the actual events are mutually exclusive. "</span>

<span>The probabilities are not mutually exclusive. Based on the group </span>

<span>P(Type O) = 9/20 = 45% or 0.45 </span>
<span>P(Type A) = 2/5 = 40% or 0.40 </span>
<span>P(Other) = 3/20 = 15% or 0.15
I hope my answer helped you. Feel free to ask more questions. Have a nice day!</span>
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The graph of which function has an axis of symmetry at x =-1/4 ?
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we know that

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The axis of symmetry is equal to the x-coordinate of the vertex

so

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we will determine in each case the axis of symmetry to determine the solution

<u>case A)</u> f(x)=2x^{2}+x-1

<u>Convert to vertex form</u>

Group terms that contain the same variable, and move the constant to the opposite side of the equation

f(x)+1=2x^{2}+x

Factor the leading coefficient

f(x)+1=2(x^{2}+0.5x)

Complete the square. Remember to balance the equation by adding the same constants to each side

f(x)+1+0.125=2(x^{2}+0.5x+0.0625)

f(x)+1.125=2(x^{2}+0.5x+0.0625)

Rewrite as perfect squares

f(x)+1.125=2(x+0.25)^{2}

f(x)=2(x+0.25)^{2}-1.125

the vertex is the point (-0.25,-1.125)

the axis of symmetry is

x=-0.25=-\frac{1}{4}

therefore

the function f(x)=2x^{2}+x-1 has an axis of symmetry at x=-\frac{1}{4}

<u>case B)</u> f(x)=2x^{2}-x+1

<u>Convert to vertex form</u>

Group terms that contain the same variable, and move the constant to the opposite side of the equation

f(x)-1=2x^{2}-x

Factor the leading coefficient

f(x)-1=2(x^{2}-0.5x)

Complete the square. Remember to balance the equation by adding the same constants to each side

f(x)-1+0.125=2(x^{2}-0.5x+0.0625)

f(x)-0.875=2(x^{2}-0.5x+0.0625)

Rewrite as perfect squares

f(x)-0.875=2(x-0.25)^{2}

f(x)=2(x-0.25)^{2}+0.875

the vertex is the point (0.25,0.875)  

the axis of symmetry is

x=0.25=\frac{1}{4}

therefore

the function f(x)=2x^{2}-x+1 does not have a symmetry axis in x=-\frac{1}{4}

<u>case C)</u> f(x)=x^{2}+2x-1

<u>Convert to vertex form</u>

Group terms that contain the same variable, and move the constant to the opposite side of the equation

f(x)+1=x^{2}+2x

Complete the square. Remember to balance the equation by adding the same constants to each side

f(x)+1+1=x^{2}+2x+1

f(x)+2=x^{2}+2x+1

Rewrite as perfect squares

f(x)+2=(x+1)^{2}

f(x)=(x+1)^{2}-2

the vertex is the point (-1,-2)  

the axis of symmetry is

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therefore

the function  f(x)=x^{2}+2x-1 does not have a symmetry axis in x=-\frac{1}{4}  

<u>case D)</u> f(x)=x^{2}-2x+1

<u>Convert to vertex form</u>

Group terms that contain the same variable, and move the constant to the opposite side of the equation

f(x)-1=x^{2}-2x

Complete the square. Remember to balance the equation by adding the same constants to each side

f(x)-1+1=x^{2}-2x+1

f(x)=x^{2}-2x+1

Rewrite as perfect squares

f(x)=(x-1)^{2}

the vertex is the point (1,0)  

the axis of symmetry is

x=1

therefore

the function  f(x)=x^{2}-2x+1 does not have a symmetry axis in x=-\frac{1}{4}

<u>the answer is</u>

f(x)=2x^{2}+x-1

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