All categories

3 years ago

(y + x) ÷ 2 + x; x=1, y=1

1 answer:

6 0

<span>(1 + 1) ÷ 2 + 1 <span>answer is 2.

Here are the steps.

1. Reduce 1 + 1 to 2 ( 2 </span>÷ 2 + 1).

2. Reduce 2 ÷ 2 to 1 ( 1 + 1).

3. So finally the answer is 2.</span>

You might be interested in

YO PLS HELP IM STUCK!

lisov135 [29]

Answer:

thats what she said

Step-by-step explanation:

5 0

3 years ago

Required information Skip to question A die (six faces) has the number 1 painted on two of its faces, the number 2 painted on th

grigory [225]

Answer:

The change to the face 3 affects the value of P(Odd Number)

Step-by-step explanation:

Analysing the question one statement at a time.

Before the face with 3 is loaded to be twice likely to come up.

The sample space is:

$S = \{1,1,2,2,2,3\}$

And the probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{6}$

$P(1) = \frac{1}{3}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{6}$

$P(2) = \frac{1}{2}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{6}$

P(Odd Number) is then calculated as:

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{1}{3} + \frac{1}{6}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{6}$

$P(Odd\ Number) = \frac{3}{6}$

$P(Odd\ Number) = \frac{1}{2}$

After the face with 3 is loaded to be twice likely to come up.

The sample space becomes:

$S = \{1,1,2,2,2,3,3\}$

The probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{7}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{7}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{7}$

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{2}{7} + \frac{1}{7}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{7}$

$P(Odd\ Number) = \frac{3}{7}$

Comparing P(Odd Number) before and after

$P(Odd\ Number) = \frac{1}{2}$ --- Before

$P(Odd\ Number) = \frac{3}{7}$ --- After

<em>We can conclude that the change to the face 3 affects the value of P(Odd Number)</em>

7 0

3 years ago

How to factor 4(x+5)^3(x-1)^2-(x+5)^4 • 2 (x-1) by grouping?

nata0808 [166]

$\bf 4(x+5)^3(x-1)^2-(x+5)^4\cdot 2(x-1)
\\\\\\
4(x+5)^3(x-1)^2-2(x-1)(x+5)^4
\\\\\\ \underline{2}\cdot 2\underline{(x+5)^3}(x-1)\underline{(x-1)}~-~\underline{2}\underline{(x-1)}(x+5)\underline{(x+5)^3}\leftarrow 
\begin{array}{llll}
notice~the\\
common\\ \underline{factors}
\end{array}
\\\\\\
2(x+5)^3(x-1)~[~2(x-1)-(x+5)~]
\\\\\\
2(x+5)^3(x-1)~[2x-2-x-5]\implies 2(x+5)^3(x-1)(x-7)$