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4 years ago

Help me with this I don't get it help4help?

Mathematics

1 answer:

Aleks [24]4 years ago

6 0

1) Divide 3/4 by 1/2. Remember how to divide fractions. Switch the problem into a multiplication problem.
3/4 x 2/1 = 3/2 (so you could only have 1 and 1/2 servings with 1/2 pound of chicken in each serving. 

2) Set up a little algebra problem. Let x = the multiplier of the original 3/4 pounds of chicken needed to make two full servings with 1/2 pound of chicken in each. 
(3/4)x = 2(1/2) (Found this online V(0-0)v)

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The height of Mountain P is 1,086 feet.The height of Mountain Q is 4 times the height of Mountain P.The area model shown below r

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Answer:

A=4000, B=80, C=24

Step-by-step explanation:

You forgot to put the correct area model, I attached it to the answer.

We have the fact that Mountain Q is 4 times the height of Mountain P. That's the "4" we have in the left side of our model. It's like having a multiplication table, next to the "4" we have "A" and upper this we have "1000", the only thing we have to do is multiplify 4*1000=4000. The next letter we have is B and below it we have "320", we divided it by 4, 320/4=80. The last letter we have is C, and is below a "6", we only have to multiplify it by 4, 6*4=24.

At the end we only sum our

A + 320 + c = 4344 (4 times the height of Mountain P).
1000 + B + 6 = 1086(the height of the Mountain P).

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3 years ago

Let X be from a geometric distribution with probability of success p. Given that P(X > y) = (1 ???? p)y for any positive inte

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Full Question

Let X be from a geometric distribution with probability of success p.

Given that P ( X > a + b | X > a ) = q ^ { b } = P ( X > b ) for any positive integer x.

Show that for positive integers a and b, P(X > a + b | X > a) = P(X > b).

Answer:

P(X > a + b | X > a) = P(X > b)

Proved --- See Explanation

Step-by-step explanation:

Given

P ( X > a + b | X > a ) = q ^ { b } = P ( X > b )

From the above.

We can derive the following

P(X > a), P(X > b) and P(X > a + b)

P(X > a) = q^a

P(X > b) = q^b

P(X > a + b) = q^(a + b)

Using the definition of conditional probability

P(X > a + b | X > a) can be represented by P(X > a + b and X > a)/ P(X>a)

X>a+b and X>a is equivalent to X>a+b since a+b is larger than a

So,

P(X > a + b and X > a)/ P(X>a) can be rewritten as

P(X>a + b)/P(X > a)

Bringing both sides together, we're left with

P(X > a + b | X > a) = P(X>a + b)/P(X > a)

By substituton

P(X > a + b | X > a) = q^(a+b)/q^a

P(X > a + b | X > a) = q^(a + b - a)

P(X > a + b | X > a) = q^(a - a + b)

P(X > a + b | X > a) = q^b

Since P(X > b) = q^b

So,

P(X > a + b | X > a) = P(X > b)

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