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3 years ago

4+3(3x + 1) = 29 - 2x

Mathematics

2 answers:

Mademuasel [1]3 years ago

6 0

Answer:

Step-by-step explanation:

4 + 9x + 3 = 29 - 2x

9x + 7 = 29 - 2x

11x = 22

x = 2

puteri [66]3 years ago

5 0

Answer:

x=2

Step-by-step explanation:

4 + 3(3x+1)=29 − 2x

4 + (3)(3x) + (3)(1)=29 + −2x

4 + 9x + 3=29 + −2x

9x + 7=−2x + 29

9x + 7 + 2x=−2x + 29 + 2x

11x + 7=29

11x+7 − 7=29 − 7

11x=22

11x /11=22/11

x=2

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Answer:

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Step-by-step explanation:

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3 years ago

An individual repeatedly attempts to pass a driving test. Suppose that the probability of passing the test with each attempt is

vladimir1956 [14]

Answer:

a) Our random variable X="number of tests taken until the individual passes" follows a geomteric distribution with probability of success p=0.25

For this case the probability mass function would be given by:

$P(X= k) = (1-p)^{k-1} p , k = 1,2,3,...$

b) $P(X \leq 3) = P(X=1) +P(X=2) +P(X=3)$

$P(X= 1) = (1-0.25)^{1-1} *0.25 = 0.25$

$P(X= 2) = (1-0.25)^{2-1} *0.25 = 0.1875$

$P(X= 3) = (1-0.25)^{3-1} *0.25 = 0.1406$

And adding the values we got:

$P(X \leq 3) =0.25+0.1875+0.1406=0.578$

c) $P(X \geq 5) = 1-P(X$

And we can find the individual probabilities:

$P(X= 1) = (1-0.25)^{1-1} *0.25 = 0.25$

$P(X= 2) = (1-0.25)^{2-1} *0.25 = 0.1875$

$P(X= 3) = (1-0.25)^{3-1} *0.25 = 0.1406$

$P(X= 4) = (1-0.25)^{4-1} *0.25 = 0.1055$

$P(X \geq 5) = 1-[0.25+0.1875+0.1406+0.1055]= 0.316$

Step-by-step explanation:

Previous concepts

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

$P(X=x)=(1-p)^{x-1} p$

Let X the random variable that measures the number of trials until the first success, we know that X follows this distribution:

$X\sim Geo (1-p)$

Part a

Our random variable X="number of tests taken until the individual passes" follows a geomteric distribution with probability of success p=0.25

For this case the probability mass function would be given by:

$P(X= k) = (1-p)^{k-1} p , k = 1,2,3,...$

Part b

We want this probability:

$P(X \leq 3) = P(X=1) +P(X=2) +P(X=3)$

We find the individual probabilities like this:

$P(X= 1) = (1-0.25)^{1-1} *0.25 = 0.25$

$P(X= 2) = (1-0.25)^{2-1} *0.25 = 0.1875$

$P(X= 3) = (1-0.25)^{3-1} *0.25 = 0.1406$

And adding the values we got:

$P(X \leq 3) =0.25+0.1875+0.1406=0.578$

Part c

For this case we want this probability:

$P(X \geq 5)$

And we can use the complement rule like this:

$P(X \geq 5) = 1-P(X$

And we can find the individual probabilities:

$P(X= 1) = (1-0.25)^{1-1} *0.25 = 0.25$

$P(X= 2) = (1-0.25)^{2-1} *0.25 = 0.1875$

$P(X= 3) = (1-0.25)^{3-1} *0.25 = 0.1406$

$P(X= 4) = (1-0.25)^{4-1} *0.25 = 0.1055$

$P(X \geq 5) = 1-[0.25+0.1875+0.1406+0.1055]= 0.316$

3 0

3 years ago

1/2 divided by 2 equal to?

valkas [14]

Answer:

.25

Step-by-step explanation:

1/2 / 2 = .25

A like would help me :)

5 0

3 years ago

Read 2 more answers

Five bells begin to ring together and they ring at intervals of 3, 6, 10, 12 and 15 seconds, respectively. How many times will t

Evgen [1.6K]

Answer:

60 times will they ring together at the same second in one hour excluding the one at the end.

Step-by-step explanation:

Given : Five bells begin to ring together and they ring at intervals of 3, 6, 10, 12 and 15 seconds, respectively.

To find : How many times will they ring together at the same second in one hour excluding the one at the end?

Solution :

First we find the LCM of 3, 6, 10, 12 and 15.

2 | 3 6 10 12 15

2 | 3 3 5 6 15

3 | 3 3 5 3 15

5 | 1 1 5 1 5

| 1 1 1 1 1

$LCM(3, 6, 10, 12,15)=2\times 2\times 3\times 5$

$LCM(3, 6, 10, 12,15)=60$

So, the bells will ring together after every 60 seconds i.e. 1 minutes.

i.e. in 1 minute they rand together 1 time.

We know, 1 hour = 60 minutes

So, in 60 minute they rang together 60 times.

Therefore, 60 times will they ring together at the same second in one hour excluding the one at the end.

6 0

3 years ago

Please help me its about solving system by substitution

Andrew [12]

Answer:

(-3, 4)

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtract Property of Equality

Algebra I

Terms/Coefficients
Solving systems of equations using substitution/elimination

Step-by-step explanation:

Step 1: Define Systems

y = -x + 1

2x + 3y = 6

Step 2: Solve for x

Substitution

Substitute in y: 2x + 3(-x + 1) = 6
Distribute 3: 2x - 3x + 3 = 6
Combine like terms: -x + 3 = 6
Isolate x terms: -x = 3
Isolate x: x = -3

Step 3: Solve for y

Define equation: y = -x + 1
Substitute in x: y = -(-3) + 1
Simplify: y = 3 + 1
Add: y = 4

6 0

3 years ago