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telo118 [61]
2 years ago
9

in the voting for city council precinct 5,only 1/2 of all eligible voters cast votes. What fraction of all eligible voters voted

for Shelley? Morgan? who received the most votes ? Shelley 3/10 Morgan 5/8
Mathematics
1 answer:
balandron [24]2 years ago
3 0
Morgan, because 5/8 *2 = 10/16 and shell's is 6/20
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276 - 35d

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-103/210

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11ピースのスライスチーズが入ったパッケージの価格は2.97ドルです。<br><br> 18スライスのパッケージは、スライスあたり同じ価格でいくらかかりますか?あなたの理由を説明または示してください。
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A regional automobile dealership sent out fliers to prospective customers indicating that they had already won one of three diff
igor_vitrenko [27]

Answer:

(a) 31101

(b) $5.65

(c) $113.38

Step-by-step explanation:

The complete question is:

A regional automobile dealership sent out fliers to prospective customers indicating that they had already won one of three different prizes: an automobile valued at $20,000, a $125 gas card, or a $5 shopping card. To claim his or her prize, a prospective print on the back of the flier listed the probabilities of winning. The chance of winning the car was 1 out of 31,101, the chance of winning the gas card was 1 out of 31,101, and the chance of winning the shopping card was 31, 099 out of 31,101.

Solution:

The information provided is as follows:

P(\text{Car})=\frac{1}{31101}\\\\P(\text{Gas Card})=\frac{1}{31101}\\\\P(\text{Shopping Card})=\frac{31099}{31101}

(a)

The number of fliers the automobile dealership sent​ out is, <em>n</em> = 31,101.

This is because the probability of winning any of the three prize is out of 31,101.

(b)

Compute the expected value of the prize won by a prospective customer receiving a​ flier as follows:

E(X)=\sum x\times P(X=x)

         =[20000\times \frac{1}{31101}]+[125\times \frac{1}{31101}]+[5\times \frac{31099}{31101}]\\\\=0.6431+0.0040+4.9997\\\\=5.6468\\\\\approx 5.65

Thus, the expected value of the prize won by a prospective customer receiving a​ flier is $5.65.

(c)

Compute the standard deviation of the prize won by a prospective customer receiving a​ flier as follows:

SD(X)=\sqrt{V(X)}=\sqrt{E(X^{2}-(E(X))^{2}}

           =\sqrt{[(20000)^{2}\times \frac{1}{31101}+(125)^{2}\times \frac{1}{31101}+(5)^{2}\times \frac{31099}{31101}]-(5.65)^{2}}\\\\=\sqrt{12854.9011}\\\\=113.37947\\\\\approx 113.38

Thus, the standard deviation of the prize won by a prospective customer receiving a​ flier is $113.38.

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