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ycow [4]
3 years ago
5

Jeremy swims 5 3/5 kilometers in a 7 day period. He swims the same distance each day. What distance does he swim?

Mathematics
2 answers:
iogann1982 [59]3 years ago
5 0

Answer: The answer is 4/5 .

Step-by-step explanation: There's a twist it says ( Eli chose A as the correct answer . How did he get that answer ?

PIT_PIT [208]3 years ago
4 0
Since 5 3/5 can be turned into 28/5, then divided by 7/1, you will get 4/5. So Jeremy swims 4/5 of a kilometer each day.
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Which expressions are differences of squares?
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If an expression is the difference of two squares, it will follow this format:
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6 0
3 years ago
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Find the area of the shape shown below.
rodikova [14]

Answer:

8

Step-by-step explanation:

First your rectangle

A = wl

A  = 2 * 2

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Then one of your triangles

A = h b/2

A = 2 * 2/2

A = 2

you have 2 triangles, so the area is 4 + 4 from the rectangle

5 0
2 years ago
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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.

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(−6,−4) would be your answer! (:
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