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Sergio
10 days ago
6

Bart found 20 quadrilaterals in his classroom. He made a Venn diagram using the properties of the quadrilaterals, comparing thos

e with four equal side lengths (E) and those with four right angles (R).
Given that a randomly chosen quadrilateral has four right angles, what is the probability that the quadrilateral also has four equal side lengths? Express your answer in percent form, rounded to the nearest whole percent.

___%
Mathematics
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What is a34 of the sequence 9,6,3,
PIT_PIT [12445]

Step-by-step explanation:

What is a34 of the sequence 9,6,3,..

r=a2-a1

r=6-9

r=-3

a34=a1+33.r

a34=9+33.(-3)

a34= 9-99

a34= -90

hope this helps!

bye!

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2 months ago
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the diameter of a cylindrical box of CDs is 5 inches. The height of the box is 4 inches. Whuch measurement is closest to the tot
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To calculate the area, simply multiply 5 by 4, resulting in 20.
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You and a friend play a game where you each toss a balanced coin. If the upper faces on the coins are both tails, you win $1; if
Zina [12379]

Answer: The mean and variance of Y are $0.25 and $6.19 respectively.

Step-by-step explanation:

The scenario is as follows: You and a friend participate in a game involving tossing a fair coin.

The sample space for tossing two coins is {TT, HT, TH, HH}

Let Y represent the earnings from one round of the game.

If both faces are heads, you win $1; therefore, P(Y=1)=P(TT)=\dfrac{1}{4}=0.25

You win $6 if both faces are heads, so P(Y=6)=P(HH)=\dfrac{1}{4}=0.25

If the faces do not match, you lose $3 which means P(Y=1)=P(TH, HT)=\dfrac{2}{4}=0.50

To find the expected value to win: E(Y)=\sum_{i=1}^{i=3} y_ip(y_1)

=1\times0.25+6\times0.25+(-3)\times0.50=0.25

Thus, the mean of Y: E(Y)= $0.25

E(Y^2)=\sum_{i=1}^{i=3} y_i^2p(y_i)\\\\=1^2\times0.25+6^1\times0.25+(-3)^2\times0.5\\\\=0.25+1.5+4.5=6.25

Variance = E[Y^2]-E(Y)^2

=6.25-(0.25)^2=6.25-0.0625=6.1875\approx6.19

Therefore, variance of Y = $ 6.19

6 0
2 months ago
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