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Mkey [24]
3 years ago
11

Each of the 25 balls in a certain box is either red, blue, or white and has a number from 1 to 10 painted on it. If one ball is

to be selected at random from the box, what is the probability that the ball selected will either be white or have an even number painted on it?
Mathematics
1 answer:
LekaFEV [45]3 years ago
8 0

Answer: Insufficient Information

The probability of selecting a red ball, a blue ball or a white ball is unknown.

Step-by-step explanation:

The probability of selecting a red ball, a blue ball or a white ball is unknown.

There could be 13 red balls, 1 blue ball and 11 white balls in the bag. There could be 3 red balls, 2 blue balls and 20 white balls in the bag. It is impossible to say with the  information given have even numbers on them. The same thing applies to the number of balls with numbers or even numbers on them. The question does not specify how many white balls have each number on them.

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What is the result when the number 79 is decreased by 17%?
trapecia [35]

Answer:

65.57 should be the correct answer!

Step-by-step explanation:

Here is the whole work I did, hope this helps!

100 - 17 = 83

79 x 0.83 = 65.57

8 0
3 years ago
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Can anyone help with either one of these
Vlada [557]
slope-point form:
we need the slope (m) and a point.
y-y₀=m(x-x₀)
Given two point A(x₁,y₂) and B(x₂,y₂), the slope of the line  is :
m=(y₂-y₁) /(x₂-x₁)

Example 3:
we can take two points:
A(12,2)
B(13,7)
m=(7-2) / (13-12)=5/1=5

therefore:
y-2=5(x-12)
y-2=5x-60
y=5x-60+2
y=5x-58

answer: y=5x-58

Example 4:
we can take two points.
A(0,0)
B(3,1)
m=(1-0)/(3-0)=1/3

Therefore:
y-0=1/3(x-0)
y=x/3

answer: y=x/3
4 0
4 years ago
Which second-degree polynomial function has a leading coefficient of –1 and root 5 with multiplicity 2?
PIT_PIT [208]
A second degree polynomial function has the general form:

                  \displaystyle{f(x)=ax^2+bx+c, where a\neq0.

The leading coefficient is a, so we have a=-1.

5 is a double root means that :

i) f(5)=0,
ii) the discriminant D is 0, where D=b^2-4ac.

Substituting x=5, we have

                                        f(5)=a(5)^2+b(5)+c,

and since f(5)=0, and a is -1 we have:

                                          0=-25+5b+c
thus c=25-5b.


By ii) \displaystyle{b^2-4ac=0.

Substituting a with -1 and c with 25-5b we have:
                                     
          \displaystyle{b^2-4ac=0
          \displaystyle{b^2-4(-1)(25-5b)=0
          \displaystyle{b^2+4(25-5b)=0
          \displaystyle{b^2-20b+100=0
          \displaystyle{(b-10)^2=0
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Finally we find c: c=25-5b=25-50=-25

Thus the function is        \displaystyle{f(x)=-x^2+10x-25


Remark: It is also possible to solve the problem by considering the form

f(x)=-1(x-5)^2 directly.

In general, if a quadratic function has leading coefficient a, and has a root r of multiplicity 2, then its form is f(x)=a(x-r)^2
5 0
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Answer:

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maw [93]
The answer is going to be A.) -2

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4 years ago
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