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AVprozaik [17]
3 years ago
13

Anyone wanna help a high school student out?

Mathematics
1 answer:
qwelly [4]3 years ago
7 0
The answer is B, because all 4 sides should be equal on a rhombus.
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Could someone help? please the deadline is in 20 min!!!
Zepler [3.9K]

Answer:

7.$95 8. $112.50

Step-by-step explanation:

7.

50 divided by 100 is 0.5, then times by 90 is 45.

45 plus 50 is 95.

8.

$150/4=$37.50

$150-$37.50=$112.50

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hichkok12 [17]
The anwser to the problem is C.

7 0
3 years ago
Bob needs to buy dirt to fill a garden bed that is a 8 ft. by 12 ft.
kondor19780726 [428]

Answer:

576 cubic feet of dirt

Step-by-step explanation:

l x w x h

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4 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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