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Cloud [144]
3 years ago
7

A set of twenty cards is labeled 1 through 20. Anna draws one card randomly. What is the probability that Anna will draw an even

number greater than 10?
Mathematics
2 answers:
Lera25 [3.4K]3 years ago
8 0

Answer:

50% chance

Step-by-step explanation:

Alina [70]3 years ago
4 0

Answer:

1/20

Step-by-step explanation:

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HELP PLS i’ll mark as brainliest if correct
Vlada [557]

Answer:

I believe the answer is the second one

Step-by-step explanation:

7 0
3 years ago
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Triangle JKL has vertices J(2,5), K(1,1), and L(5,2). Triangle QNP has vertices Q(-4,4), N(-3,0), and P(-7,1). Is (triangle)JKL
Tems11 [23]

Answer:

Yes they are

Step-by-step explanation:

In the triangle JKL, the sides can be calculated as following:

  • J(2;5); K(1;1)

             => JK = \sqrt{(1-2)^{2} + (1-5)^{2}  } = \sqrt{(-1)^{2}+(-4)^{2}  } = \sqrt{1+16}=\sqrt{17}

  • J(2;5); L(5;2)

             => JL = \sqrt{(5-2)^{2} + (2-5)^{2}  } = \sqrt{3^{2}+(-3)^{2}  } = \sqrt{9+9}=\sqrt{18} = 3\sqrt{2}

  • K(1;1); L(5;2)

             =>  KL = \sqrt{(5-1)^{2} + (2-1)^{2}  } = \sqrt{4^{2}+1^{2}  } = \sqrt{1+16}=\sqrt{17}

In the triangle QNP, the sides can be calculate as following:

  • Q(-4;4); N(-3;0)

             => QN = \sqrt{[-3-(-4)]^{2} + (0-4)^{2}  } = \sqrt{1^{2}+(-4)^{2}  } = \sqrt{1+16}=\sqrt{17}

  • Q (-4;4); P(-7;1)

   => QP = \sqrt{[-7-(-4)]^{2} + (1-4)^{2}  } = \sqrt{(-3)^{2}+(-3)^{2}  } = \sqrt{9+9}=\sqrt{18} = 3\sqrt{2}

  • N(-3;0); P(-7;1)

             =>  NP = \sqrt{[-7-(-3)]^{2} + (1-0)^{2}  } = \sqrt{(-4)^{2}+1^{2}  } = \sqrt{16+1}=\sqrt{17}

It can be seen that QPN and JKL have: JK = QN; JL = QP; KL = NP

=> They are congruent triangles

7 0
3 years ago
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5(n+1) = 2n + 20 can someone tell me the answer and explain
finlep [7]

Answer:

n=15

Step-by-step explanation:

5(n+1)=2n+20

U start by expanding the first term which will give you 5n+5

Then after that u collect like terms giving u

5n-2n=20-5

Finally u evaluate it giving u

n=15

8 0
3 years ago
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5 0
4 years ago
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Please answer my questions is really easy but idk
GrogVix [38]

Given:

m \angle A B D=96^{\circ}

m \angle 2 =m \angle 1+ 26^{\circ}

To find:

m \angle 1

Solution:

m \angle DBC+ m \angle CBA = m \angle ABD

m \angle 1+ m \angle 2 = m \angle ABD

Substitute m \angle 2 =m \angle 1+ 26^{\circ}.

m \angle 1+ m \angle 1 + 26^\circ = 96^\circ

2m \angle 1+ 26^\circ = 96^\circ

Subtract 26° from both sides.

2m \angle 1+ 26^\circ -26^\circ = 96^\circ -26^\circ

2m \angle 1 = 70^\circ

Divide by 2 on both sides, we get

m \angle 1=35^{\circ}

Therefore, m∠1 = 35°.

8 0
4 years ago
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