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meriva
3 years ago
15

Rewrite the equation below so that it does not have fractions four -4/9 times equals two over three do not use decimals in your

answer please help me I’ll give you 10 points

Mathematics
1 answer:
Rama09 [41]3 years ago
6 0

The equation is 36 - 4x = 6.

Solution:

To write the equation without fraction.

$4-\frac{4}{9} x=\frac{2}{3}

$\frac{4}{1} -\frac{4}{9} x=\frac{2}{3}

The denominators must be same to add/subtract the fractions.

LCM of 1, 9 and 3 = 9.

Multiply first term by \frac{9}{9} and \frac{2}{3} by \frac{3}{3} to make the denominator same.

$\frac{4\times9}{1\times9} -\frac{4}{9} x=\frac{2\times3}{3\times3}

$\frac{36}{9} -\frac{4}{9} x=\frac{6}{9}

$\frac{36-4x}{9} =\frac{6}{9}

Multiply by 9 on both sides.

$\frac{36-4x}{9}\times9 =\frac{6}{9}\times9

Both 9 in the numerator and denominator get canceled.

36 - 4x = 6

This term does not have fraction.

Hence the equation is 36 - 4x = 6.

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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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