All categories

2 years ago

For which real numbers x and y is it true that x + y = x+y?

Mathematics

1 answer:

Alchen [17]2 years ago

6 0

By the general application of cumulative property of addition :

x + y = y + x

For sure

You might be interested in

Help how do I solve it

Mila [183]

When two lines intersect, opposite angles are equal. This means that angle 1 equals angle 4. We can use that information to find their values.
Angle 1 = Angle 4
6n+1 = 4n+19
2n=18
n=9

6(9)+1=54+1=55

Angle 1 and 4 equal 55 degrees.

Two angles that form a straight line together have a total sum of 180 degrees. Angles 1 and 5 are like this, as well as Angles 4 and 5, and Angles 4, 3, and 2 added together.

Therefore, 180 = (angle 4) + (angle 3) + (angle 2)
180= 55+(angle 3) + (angle 2)
125= angle 3 + angle 2

I'm not sure what else can be extrapolated from this. There doesn't seem to be a way to find out what the measure of angle 2 is without angle 3 as well. I hope this helps and you can figure it out from the answer choices!

5 0

2 years ago

A bag contains 35 marbles, 11 of which are red. A marble is randomly selected from the bag, and it is blue. This blue marble is

Ilia_Sergeevich [38]

Answer:

11 red + 24 blue = 35 marbles

If 1 blue is withdrawn

11 red + 23 blue = 34 marbles

P = 23 / 34 = .38 probability of drawing blue marble

6 0

2 years ago

The prior probabilities for events A1 and A2 are P(A1) = 0.20 and P(A2) = 0.80. It is also known that P(A1 ∩ A2) = 0. Suppose P(

Umnica [9.8K]

Answer:

(a) $A_1$ and $A_2$ are indeed mutually-exclusive.

(b) $\displaystyle P(A_1\; \cap \; B) = \frac{1}{20}$ , whereas $\displaystyle P(A_2\; \cap \; B) = \frac{1}{25}$ .

(d) $\displaystyle P(A_1 \; |\; B) \approx \frac{5}{9}$ , whereas $P(A_1 \; |\; B) = \displaystyle \frac{4}{9}$

Step-by-step explanation:

$P(A_1 \; \cap \; A_2) = 0$ means that it is impossible for events $A_1$ and $A_2$ to happen at the same time. Therefore, event $A_1$ and $A_2$ are mutually-exclusive.

By the definition of conditional probability:

$\displaystyle P(B \; | \; A_1) = \frac{P(B \; \cap \; A_1)}{P(B)} = \frac{P(A_1 \; \cap \; B)}{P(B)}$ .

Rearrange to obtain:

$\displaystyle P(A_1 \; \cap \; B) = P(B \; |\; A_1) \cdot P(A_1) = 0.25 \times 0.20 = \frac{1}{20}$ .

Similarly:

$\displaystyle P(A_2 \; \cap \; B) = P(B \; |\; A_2) \cdot P(A_2) = 0.80 \times 0.05 = \frac{1}{25}$ .

Note that:

$\begin{aligned}P(A_1 \; \cup \; A_2) &= P(A_1) + P(A_2) - P(A_1 \; \cap \; A_2) = 0.20 + 0.80 = 1\end{aligned}$ .

In other words, $A_1$ and $A_2$ are collectively-exhaustive. Since $A_1$ and $A_2$ are collectively-exhaustive and mutually-exclusive at the same time:

$\displaystyle P(B) = P(B \; \cap \; A_1) + P(B \; \cap \; A_2) = \frac{1}{20} + \frac{1}{25} = \frac{9}{100}$ .

By Bayes' Theorem:

$\begin{aligned} P(A_1 \; |\; B) &= \frac{P(B \; | \; A_1) \cdot P(A_1)}{P(B)} \\ &= \frac{0.25 \times 0.20}{9/100} = \frac{0.05 \times 100}{9} = \frac{5}{9}\end{aligned}$ .

Similarly:

$\begin{aligned} P(A_2 \; |\; B) &= \frac{P(B \; | \; A_2) \cdot P(A_2)}{P(B)} \\ &= \frac{0.05 \times 0.80}{9/100} = \frac{0.04 \times 100}{9} = \frac{4}{9}\end{aligned}$ .