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Svetradugi [14.3K]
3 years ago
6

Please help me with this one

Mathematics
1 answer:
Allushta [10]3 years ago
5 0

Answer:

248

Step-by-step explanation:

Divide 62 by 2 to get 31. Each distance on the map should be multiplied by this number. So if it's 8 on the map, its 8 x 31 in real life. 8 x 31 = 248

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April has 5 milligrams of zinc and 0.25 milligrams of copper.
GREYUIT [131]

Answer:

It's 4.75

Step-by-step explanation:

If you subtract 5-.25=? The answer would be 4.75. Hope this helps.

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2 years ago
A car traveled at an average rate of 51 miles per hour on one highway. It then traveled at an average rate of 71 miles per hour
sergiy2304 [10]

d1 = 51(t)

d2 = 71(8 - t)

d1 + d2 = 508

and now you just have to solve for t.

3 0
2 years ago
Can someone please help me on these?
JulijaS [17]
The answer is C 7v-1
7 0
3 years ago
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Anna walks her dog at a constant rate of 12 blocks in 8 minutes.​
mart [117]

Answer:

she will reach 12 blocks in 8 minutes

Step-by-step explanation:

next time actually ask a question dont just state a fact. XD

8 0
3 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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