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Allushta [10]
2 years ago
12

Ronald spins a spinner that has 10 sections numbered from 1 to 10. What is the probability that the spinner lands on a prime num

ber?
Mathematics
2 answers:
blagie [28]2 years ago
8 0

Answer:

40%

Step-by-step explanation:

Prime numbers less than 10 are 2, 3, 5, and 7.  Assuming the sections are sized equally, then four out of ten sections is 40%.

kobusy [5.1K]2 years ago
7 0

The probability the spinner lands on a prime number is 4/10 or 40%

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Which of the following are exterior angles? Check all that apply.
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An exterior angle of a triangle is an angle formed by one side of the triangle and the extension of an adjacent side of the triangle.

FACTS:

  • Every triangle has 6 exterior angles, two at each vertex.
  • Notice that the "outside" angles that are "vertical" to the angles inside the triangle are NOT called exterior angles of a triangle.

1. Angles 1, 5 and 6 are inteior angles of the given triangle.

2. Angle 3 is vertical to the interior angle 5.

3. Angles 2 and 4 are exterior angles of the triangle.

Answer: correct options are A and B.

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Mike took 12 shots with a basketball and made 9. This can be expressed as a decimal as 0.75. What other person’s successful shot
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I baseball player swung at the ball 100 times and only made conract 75 times this could be exspressed as 0.75 
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The sum of two numbers is 361 and the difference between the two number is 173 what are the two numbers
melamori03 [73]
The answer is attached

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Which number is located between 1.2 and 1.4 on the number line
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Find the equation of the plane through the point (2,5,7) that is parallel to the line r=(3i+2j−2k)+t(i+2j+9k) and perpendicular
insens350 [35]

The plane we want to find has general equation

a(x-2)+b(y-5)+c(z-7)=0

with a,b,c not equal to 0, and has normal vector

\vec n=a\,\vec\imath+b\,\vec\jmath+c\,\vec k

\vec n is perpendicular to both the normal vector of the other plane, which is 4\,\vec\imath+5\,\vec\jmath+6\,\vec k, as well as the tangent vector to the line \vec r(t), which is \vec\imath+2\,\vec\jmath+9\,\vec k.

This means the dot product of \vec n with either vector is 0, giving us

\begin{cases}4a+5b+6c=0\\a+2b+9c=0\end{cases}

Suppose we fix c=1. Then the system reduces to

\begin{cases}4a+5b=-6\\a+2b=-9\end{cases}

and we get

(4a+5b)-4(a+2b)=-6-4(-9)\implies-3b=30\implies b=-10

a+2(-10)=-9\implies a=11

Then one equation for the plane could be

\boxed{11(x-2)-10(y-5)+(z-7)=0}

or in standard form,

\boxed{11x-10y+z=-21}

The solution is unique up to non-zero scalar multiplication, which is to say that any equation (11x-10y+z)k=-21k would be a valid answer. For example, suppose we instead let c=2; then we would have found a=22 and b=-20, but clearly dividing both sides of the equation

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by 2 gives the same equation as before.

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2 years ago
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