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levacccp [35]
3 years ago
12

Help meh?

Mathematics
1 answer:
zloy xaker [14]3 years ago
3 0
X= # of miles
.55x+1.75 is greater than or equal to 10

So if she has to go more than 2 miles we will put 2 in for X

.55(2)+1.75 is greater than or equal to 10
.55×2= 1.10+1.75= 2.85
2.85 is greater than 10 is false so now we know she has to go more than 2 miles.

Let's guess and check. Now let's say 15 miles
.55×15= 8.25+1.75= 10
10=10 So she needs to drive at least 15 miles

x is greater than or equal to 15

Please don't solely rely on my answer I don't know if it's correct
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Answer:

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Step-by-step explanation:

The first trial among a distribution of independent trials with a constant success probability follows a geometric probability.

The expected value of a negative binomial is the reciprocal of its success probability <em>p.</em>

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<em>E(X1)</em>=1

After the first prize is drawn, we are interested in the first time (success), that we draw another prize that is different (<em>p</em>=9/10)

E(X2)=1/9÷10

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After the first prizes have being drawn, we now interested in the first time success that we draw a different prize from the first 2

E(X3)=1/8÷10

E(X3)=10/8

E(X3)=5/4

After the first three prizes have been drawn, we are now interested in the success of a first time draw for the 4th different prize

E(X4)=1/7÷10

E(X4)=10/7

Fifth different prize

E(X5)=1/6÷10

E(X5)=10/6

E(X5)=5/3

Sixth different prize

E(X6)=1/5÷10

E(X6)=10/5

E(X6)=2

Seventh different prize

E(X7)=1/4÷10

E(X7)=10/4

E(X7)=5/2

Eighth different prize

E(X8)=1/3÷10

E(X8)=10/3

Ninth different prize

E(X9)=1/2÷10

E(X9)=10/2

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After the 9 different prizes have been drawn, we are interested in the first time we will draw the tenth different prize

E(X10)=1/1÷10

E(X10)=10

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<em>E(X)=</em>E(X1)+E(X2)+E(X3)+E(X4)+E(X5)+E(X6)+E(X7)+E(X8)+E(X9)+E(X10)

<em>E(X)=</em>1+10/9+5/4+10/7+5/3+2+5/2+10/3+5+10

<em>E(X)</em>= 29.29

Note: Those values in <em>E(X), </em>should be written the way it will in standard algebra ie they should be small.

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