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3 years ago

How do you use the associative property to write an expression equivalent to (w+9)+3

Mathematics

1 answer:

m_a_m_a [10]3 years ago

5 0

The equivalent expression for the given expression (w+9)+3 using associative property is w + 12

<u>Solution:</u>

Given that, expression is ( w + 9 ) + 3

We have to find how do we can use the associative property to write an expression equivalent to ( w + 9 ) + 3

The associative property states that you can add or multiply regardless of how the numbers are grouped

By above definition, we get

(a + b) + c = a + (b + c)

So, now let us apply the above property to the given expression,

Then, ( w + 9 ) + 3 = w + ( 9 + 3 ) = w + 12

Hence, the equivalent expression for the given expression is w + 12

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Let X equal the number of hours that a randomly selected person from City A reads the news online each day. Suppose that X has a

Nina [5.8K]

Answer:

$P(Y>X) = \frac{17}{32}$

Step-by-step explanation:

Recall that since X is uniformly distributed over the set [1,4] we have that the pdf of X is given by $f(x) = \frac{1}{3}$ if $1\leq x \leq 4$ and 0 otherwise. In the same manner, the pdf of Y is given by $g(y) = \frac{1}{4}$ if $1\leq y\leq 5$ and 0 otherwise.

Note that if Y is in the interval (4,5] then Y>X by default. So, in this case we have that P(Y>X| y in (4,5]) = 1. We want to calculate the probability of having Y in that interval . That is

$P(Y>4) = \int_{4}^{5}\frac{1}{4}dy = \frac{1}{4}$ . Thus, $P(Y\leq 4 ) = 1 - P(Y>4)= \frac{3}{4}$ .

We want to proceed as follows. Using the total probability theorem, given two events A, B we have that

$P(A) = P(A|B)\cdot P(B) + P(A|B^c) \cdot P(B^c)$ In this case, A is the event that Y>X and B is the event that Y is in the interval (4,5].

If we assume that X and Y are independent, then we have that the joint pdf of X,Y is given by $h(x,y) = f(x)g(y) = \frac{1}{12}$ when $1\leq x \leq 4, 1\leq y \leq 4$ . We can draw the region were Y>X and the function h(x,y) is different from 0. (The drawing is attached). This region is described as follows: $1\leq x \leq 4$ and $x\leq y \leq 4$ , then (the specifics of the calculations of the integrals are ommitted)

$P(Y>X| y \in [1,4)) = \int_{1}^{4}\int_{x}^{4}\frac{1}{12}dy dx = \frac{1}{12}\int_{1}^{4} (4-x) dx = \frac{1}{12}\left.(4x-\frac{x}{2})\right|_{1}^{4}= \frac{1}{12}(4\cdot 4 - \frac{4^2}{2}-(4-\frac{1}{2}) = \frac{9}{2\cdot 12} = \frac{3}{8}$

Thus,

$P(Y>X) = 1\cdot \frac{1}{4} + \frac{3}{8}\cdot \frac{3}{4} = \frac{17}{32}$

5 0

4 years ago

!PLEASE HELP ME!

Mkey [24]

Answer:

We conclude that Tim is correct when he says that the expression x² only yields values that are positive.

Step-by-step explanation:

Given the expression

x²

Plug in and checking x = 1 and x = -1 in the expression

Putting x = 1 in the expression

x²= (1)² = 1

Putting x = -1 in the expression

x²= (-1)² = 1

Thus, the expression yields the same output '1' when we enter x = 1, and x=-1.

Plug in and checking x = 2 and x = -2 in the expression

Putting x = 2 in the expression

x²= (2)² = 4

Putting x = -1 in the expression

x²= (-2)² = 4

Thus, the expression yields the same output '4' when we enter x = 2, and x=-2.

Plug in and checking x = 3 and x = -3 in the expression

Putting x = 3 in the expression

x²= (3)² = 9

Putting x = -1 in the expression

x²= (-3)² = 9

Thus, the expression yields the same output '9' when we enter x = 3, and x=-3.

The reason why the expression x² only yields positive values because the expression is in the square form, and the square of any number will always yield a positive value, no matter whether the input number is negative or positive.

Therefore, we conclude that Tim is correct when he says that the expression x² only yields values that are positive.

3 0

3 years ago