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IrinaVladis [17]
3 years ago
11

A 3-column table with 4 rows. The first column is labeled Currency and entries Canadian dollar, Euro, Japanese yen, Indian rupee

. The second column is labeled U S D/1 Unit and entries 0.97071, 1.35261, 0.01007, 0.01594. The third column is labeled Units/1 U S D and entries 1.03069, 0.73935, 99.90487, 62.72053.
Mitchell travels from the US to Canada, where he exchanges 150 US dollars for Canadian dollars. He then spends 20 Canadian dollars, returns to the US, and exchanges the remaining money back to US dollars. How many US dollars does Mitchell have remaining
Mathematics
2 answers:
MAXImum [283]3 years ago
6 1

Answer:

Its A. just took the test

Step-by-step explanation:

kherson [118]3 years ago
5 0

Answer:

A. 129.46

B. 130.00

C. 130.66

D. 134.56

The correct answer is c. 130.66

Step-by-step explanation:

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8 0
2 years ago
A player tosses a die 6 times.If gets a number 6 Atleast two times he wins 2 dollars ,otherwise he looses 1 dollar.. Find the ex
swat32

Answer:

E(x)=-0.2101

Step-by-step explanation:

The expected value for a discrete variable is calculated as:

E(x)=x_1*p(x_1)+x_2*p(x_2)

Where x_1 and x_2 are the values that the variable can take and p(x_1) and p(x_2) are their respective probabilities.

So, a player can win 2 dollars or looses 1 dollar, it means that x_1 is equal to 2 and x_2 is equal to -1.

Then, we need to calculated the probability that the player win 2 dollars and the probability that the player loses 1 dollar.

If there are n identical and independent events with a probability p of success and a probability (1-p) of fail, the probability that a events from the n are success are equal to:

P(a)=nCa*p^a*(1-p)^{n-a}

Where nCa=\frac{n!}{a!(n-a)!}

So, in this case, n is number of times that the player tosses a die and p is the probability to get a 6. n is equal to 6 and p is equal to 1/6.

Therefore, the probability  p(x_1) that a player get at least two times number 6, is calculated as:

p(x_1)=P(x\geq2)=1-P(0)-P(1) \\\\P(0) =6C0*(1/6)^{0}*(5/6)^{6}=0.3349\\P(1)=6C1*(1/6)^{1}*(5/6)^{5}=0.4018\\\\p(x_1)=1-0.3349-0.4018\\p(x_1)=0.2633

On the other hand, the probability p(x_2) that a player don't get  at least two times number 6, is calculated as:

p(x_2)=1-p(x_1)=1-0.2633=0.7367

Finally, the expected value of the amount that the player wins is:

E(x)=x_1*p(x_1)+x_2*p(x_2)\\E(x)=2*(0.2633)+ (-1)*0.7367\\E(x)=-0.2101

It means that he can expect to loses 0.2101 dollars.

3 0
3 years ago
Read 2 more answers
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