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3 years ago

How many solutions does the equation have? 2x + 3 = 4x + 2

2 answers:

8 0

Well, solve for x.

Combine like terms by performing the opposite operation of subtracting 4x on both sides of the equation

The 4x's will cross out on the right

4x - 4x = 0x = 0

On the left:

2x - 4x = -2x

Now the equation looks like:

-2x + 3 = 2

Continue to combine like terms by subtracting 3 on both sides of the equation

On the left:

3 - 3 = 0

On the right:

2 - 3 = -1

Equation:

-2x = -1

Isolate x by performing the opposite operation of dividing -2 on both sides of the equation

On the left:

-2x ÷ -2 = 1

On the right:

-1 ÷ -2 = 1/2

x= 1/2

So, there is only one solution: 1/2

pishuonlain [190]3 years ago

7 0

It is linear equation so it has only one solution.
so number of solution depends upon max degree of variable

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17% of students in a class have brown hair. If 6 students are chosen at random, what is the probability that exactly 3 have brow

dezoksy [38]

Answer:

6.8%

Step-by-step explanation:

the question has no information about class size (no. of student in class)

assume no. of student is infinite

P(3brown out of 6) = 6C3 * P(brown)^3 * P(not brown)^3

= 20 * 0.004913 * 0.695789

= 0.06837

3 0

2 years ago

Help please asap today is the deadline

Korvikt [17]

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7 0

3 years ago

A box contains 3 coins. One coin has 2 heads and the other two are fair. A coin is chosen at random from the box and flipped. If

Blababa [14]

Answer: Our required probability is $\dfrac{1}{2}$

Step-by-step explanation:

Since we have given that

Number of coins = 3

Number of coin has 2 heads = 1

Number of fair coins = 2

Probability of getting one of the coin among 3 = $\dfrac{1}{3}$

So, Probability of getting head from fair coin = $\dfrac{1}{2}$

Probability of getting head from baised coin = 1

Using "Bayes theorem" we will find the probability that it is the two headed coin is given by

$\dfrac{\dfrac{1}{3}\times 1}{\dfrac{1}{3}\times \dfrac{1}{2}+\dfrac{1}{3}\times \dfrac{1}{2}+\dfrac{1}{3}\times 1}\\\\=\dfrac{\dfrac{1}{3}}{\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{3}}\\\\=\dfrac{\dfrac{1}{3}}{\dfrac{2}{3}}\\\\=\dfrac{1}{2}$

Hence, our required probability is $\dfrac{1}{2}$

No, the answer is not $\dfrac{1}{3}$

5 0

3 years ago

Juan has 20 books to sell. He sells the books for $15 each.

Anastaziya [24]

Answer:

Well u would multiply 20 x 15 to find out the highest if the range which is 300 . So that knocks it down to B or D. But since they're $15 each, It would be based of the multiples of 15 making your answer B .

The range is all the y values.....or the f(x) values .

So the practical range would be the multiples of 15 between 0 and 300,Inclusive .

<h2>Hope this helps you!! </h2>

5 0

2 years ago

Is EFGH below a rectangle

poizon [28]

Answer:

D. No, because EFGH is a parallelogram but it’s diagonals are not congruent

Step-by-step explanation:

The differences between the end points of the diagonals are ...

F -H = (1, 1) -(2, -5) = (1 -2, 1 -(-5)) = (-1, 6)

G -E = (4, -2) -(-1, -2) = (4 -(-1), -2 -(-2)) = (5, 0)

The length of FH is more than 6, the length of GE is exactly 5. The diagonals are different length, so the figure cannot be a rectangle.

The midpoints of the diagonals will be in the same place if the sum of their end points is the same. (Dividing each sum by 2 gives the midpoint of that segment.)

F+H = (1, 1) +(2, -5) = (3, -4)

G+E = (4, -2) +(-1, -2) = (3, -4)

The diagonals bisect each other (have the same midpoint), so the figure is a parallelogram.

EFGH is a parallelogram, but not a rectangle: its diagonals are not congruent.

3 0

3 years ago