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weqwewe [10]
3 years ago
9

Solve for x and y x-y=11 2x+y=19

Mathematics
2 answers:
Ilia_Sergeevich [38]3 years ago
7 0
X=10 y=-1
U can use Photomath for questions like this
Charra [1.4K]3 years ago
6 0

Answer:

x = 10

y = -1

Step-by-step explanation:

System of equations

x - y = 11

2x + y = 19

What is x and y?

To solve this, we need to find a way we can substitute one equation into another. Let's try the first equation.

x - y = 11

Add y to both sides :

x = 11 + y

Now substitute x from the first equation into x in the second equation :

2(11 + y) + y = 19

Distribute :

22 + 2y + y = 19

Add like terms :

22 + 3y = 19

Get y alone :

Subtract 22 from both sides :

3y = -3

Divide 3 from both sides :

y = -1

Now that we have y, let's substitute it into the easier equation, which would be the first one.

x - (-1) = 11

x + 1 = 11

Subtract 1 from both sides :

x = 10

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Sergeeva-Olga [200]

Answer:

There are many ways to divided 8 shown below...

Step-by-step explanation:

8 divided by 1 = 8

8 divided by 2 = 4

8 divided by 8 = 0

4 0
3 years ago
the name Joe is very common at a school in one out of every ten students go by the name. If there are 15 students in one class,
kumpel [21]

Using the binomial distribution, it is found that there is a 0.7941 = 79.41% probability that at least one of them is named Joe.

For each student, there are only two possible outcomes, either they are named Joe, or they are not. The probability of a student being named Joe is independent of any other student, hence, the <em>binomial distribution</em> is used to solve this question.

<h3>Binomial probability distribution </h3>

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

  • One in ten students are named Joe, hence p = \frac{1}{10} = 0.1.
  • There are 15 students in the class, hence n = 15.

The probability that at least one of them is named Joe is:

P(X \geq 1) = 1 - P(X = 0)

In which:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{15,0}.(0.1)^{0}.(0.9)^{15} = 0.2059

Then:

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.2059 = 0.7941

0.7941 = 79.41% probability that at least one of them is named Joe.

To learn more about the binomial distribution, you can take a look at brainly.com/question/24863377

8 0
3 years ago
can someone please answer questions 1,3,7,9 and 10 for me im dumb and suck so bad at geometry thank you
adell [148]

Answer:1,5

Step-by-step explanation:

7 0
3 years ago
A delivery truck is transporting boxes of two sizes: large and small. the combined weight of a large box and a small box is 60 p
sdas [7]

Let a = weight of large box in pounds

Let b = weight of small box in pounds

(1) 65a + 70b=4000

(2) a+b=60

Now we can simultaneously solve these equations.

a = 60 - b. Hence 65(60 - b) + 70b = 4000 So 3900 + 5b = 4000 and 100 = 5b hence b = 20.

SImilarly a = 40. Now we can verify too by putting in the calculated values.

5 0
4 years ago
How do we solve for x
blondinia [14]
X/2 + 5 = -10
x/2 = -10 - 5
x/2 = -15
x = -15(2)
x = -30
3 0
3 years ago
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