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mel-nik [20]
3 years ago
8

What's the rule about adding intergers?​

Mathematics
1 answer:
vivado [14]3 years ago
5 0

Answer:

Integers include both positive and negative numbers, and there are several rules for adding integers. Adding two positive integers together will always result in a positive integer. For example, 6 + 4 = 10. To add two negative integers, add the numbers together and place a negative sign in front of the answer.

You might be interested in
Devaughn is 15 years younger than Sydney. The sum of their ages is 73. What is Sydney’s age?
Olegator [25]

Answer:

Sydney is 44 years old

Step-by-step explanation:

<u>Step 1:  Make a system of equations </u>

d = s - 15

<em>d + s = 73 </em>

<u>Step 2:  Plug in s - 15 for d in the second equation and solve </u>

s - 15 + s = 73

2s - 15 = 73 + 15

2s  / 2= 88 / 2

<em>s = 44 </em>

Answer:  Sydney is 44 years old

6 0
3 years ago
Read 2 more answers
You are at a European beach with 60 other visitors. 36 of them speak English. If you randomly meet two people on the beach, what
Kaylis [27]

Answer:

Assuming I'm one of the 36 English speakers and the other 24 speak Spanish for illustration purposes.  The problem can be modeled as 59 marbles with 35 E and 24 S marbles as N = 59 possible outcomes = n(E) + n(S) = 35 + 24.

So I reach into the pile of marbles (on the beach) and the probability that it's p(E) = n(E)/N = 35/59 = 0.593220339 when I meet the one person. ANS

I assume I remember that first person; so I remove him from the marbles (by avoiding him on the beach) and now my probability is p(E and E) = n(E)/N * n(E)-1/(N - 1) = 35/59*34/58 = 0.347749854 ANS

Following the same logic p(E and E and E) = 35/59*34/58*33/57 = 0.201328863 ANS

This last one is different from the first three.  This one is p(E >= 1|4 attempts).  We can trace out a probability tree to identify those branches that contain at least one E event.  So:

EEEE p() = 35/59 * 34/58 * 33/57 * 32/56 =  

EEES p() = 35/59 * 34/58 * 33/57 * 24/56 =

EESE p() = 35/59 * 34/58 * 24/57 * 33/56 =

ESEE p() = 35/59 * 24/58 * 34/57 * 33/56 =

SEEE p() = 24/59 * 35/58 * 34/57 * 33/56 =

EESS p() = 35/59 * 34/58 * 24/57 * 23/56 =

ESES p() = 35/59 * 24/58 * 34/57 * 23/56 =

SEES

SESE

SSEE

ESSS  And so on, but...a big BUT...why do all this when

SESS

SSES

SSSE

SSSS

p(E>=1|4) = 1 - p(S and S and S and S) = 1 - 24/59 * 23/58 * 22/57 * 21/56 = 0.976652619   ANS.  In other words we find the probability of not meeting an Englishman and take 1 minus that value to find the probability of meeting at least one.

00

Step-by-step explanation:

3 0
2 years ago
Please I’ve attached the question can you help me
Andreyy89

Answer:

1. <em>n</em>[(A U B) - C] = {23, 24, 27, 29, 33, 36, 37, 39, 41, 42, 43, 45, 47, 48, 51, 53, 54, 57, 59, 61, 63}

2. <em>n</em>[(A - B) U C] = <em>n</em>[A U C] = {6, 10, 12, 15, 20, 23, 29, 30, 31, 37, 41, 43, 47, 53, 59, 60, 61}

3. D. I, II and III.

Step-by-step explanation:

U = {21, 22, 23, ..., 64}

A prime number is a number that can be divided only by 1 and itself.

A = {23, 29, 31, 37, 41, 43, 47, 53, 59, 61}

B = {24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63}

C = {6, 10, 12, 15, 20, 30, 60}

1. Find <em>n</em>[(A U B) - C]

<em>n</em>(A U B) = {23, 24, 27, 29, 30, 33, 36, 37, 39, 41, 42, 43, 45, 47, 48, 51, 53, 54, 57, 59, 60, 61, 63}

<em>n</em>[(A U B) - C] = {23, 24, 27, 29, 30, 33, 36, 37, 39, 41, 42, 43, 45, 47, 48, 51, 53, 54, 57, 59, 60, 61, 63} - {6, 10, 12, 15, 20, 30, 60}

Since only 30 and 60 are common to <em>n</em>(A U B) and C, we therefore remove them it and have:

<em>n</em>[(A U B) - C] = {23, 24, 27, 29, 33, 36, 37, 39, 41, 42, 43, 45, 47, 48, 51, 53, 54, 57, 59, 61, 63}

2. Find <em>n</em>[(A - B) U C]

To get <em>n</em>(A - B) we remove all the elements of B in A. Since there are no common elements between A and B, we therefore have:

A - B = A = {23, 29, 31, 37, 41, 43, 47, 53, 59, 61}

C = {6, 10, 12, 15, 20, 30, 60}

Therefore, we have:

<em>n</em>[(A - B) U C] = <em>n</em>[A U C] = {6, 10, 12, 15, 20, 23, 29, 30, 31, 37, 41, 43, 47, 53, 59, 60, 61}

3. Which of the following is/are true?

I. A ∩ B = A ∩ C

A ∩ B = ∅

A ∩ C = ∅

Therefore, A ∩ B = A ∩ C is true.

II. A - B = A - C

A - B = A = {23, 29, 31, 37, 41, 43, 47, 53, 59, 61}

A - C = A = {23, 29, 31, 37, 41, 43, 47, 53, 59, 61}

Therefore, A - B = A - C is true.

III. A ∩ (B ∪ C) = ∅

A = {23, 29, 31, 37, 41, 43, 47, 53, 59, 61}

B U C = {6, 10, 12, 15, 20, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63}

Therefore, A ∩ (B ∪ C) = ∅ is true.

Therefore, the correct option is D i.e. I, II and III are true.

4 0
3 years ago
Evaluate 6ᵗʷᵒ – (9 ÷ x) when x = 3.​
tia_tia [17]

Answer:

B. 33

Step-by-step explanation:

36 - (9/ 3)

36 - 3

33

7 0
2 years ago
Find the solution of the system x′=−4y,y′=−2x, where primes indicate derivatives with respect to t, that satisfies the initial c
Andreas93 [3]

Answer:2y^2 = x^2+4

Step-by-step explanation:

given is a differntial equation in parametric form as

x'=-4y\\y'=-2x

We can find dy/dx by dividing y' by x'

\frac{dy}{dx} =\frac{-2x}{-4y} \\4ydy =2x dx\\2y^2 = x^2+C

(By separating the variables and integrating)

C is arbitrary constant

When x=-2, y =-2

Substitute to get

8=4+C\\C=4

So solutoin is

2y^2 = x^2+4

is the solution

5 0
3 years ago
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