-5(x + 1) s 30. What is a possible value of x?<br> help me pls

The numerator and denominator of a fraction are in the ratio of 3 to 5. If the numerator and denominator are both increased by 2

ivolga24 [154]

Question:

The numerator and denominator of a fraction are in the ratio of 3 to 5. If the numerator and denominator are both decreased by 2, the fraction is now equal to $\frac{1}{2}$ .

If n = the numerator and d = the denominator, which of the following systems of equations could be used to solve the problem?

5n = 3d and n - 2 = 2d - 4

5n = 3d and 2n - 4 = d - 2

3n = 5d and 2n - 4 = d - 2

Answer:

5n = 3d and 2n – 4 = d – 2

Solution:

Let n be the numerator of the fraction and d be the denominator of the fraction.

Given the numerator and denominator of a fraction are in the ratio of 3 to 5.

This can be written as n : d = 3 : 5.

⇒ $\frac{n}{d}= \frac{3}{5}$ – – – – (1)

Do cross multiplication, we get

⇒ 5n = 3d

When the numerator and denominator are decreased by 2, the fraction is equal to $\frac{1}{2}$ .

⇒ $\frac{n-2}{d-2}= \frac{1}{2}$

Do cross multiplication, we get

⇒ 2(n –2)=1(d – 2)

⇒ 2n – 4 = d – 2

Hence, 5n = 3d and 2n – 4 = d – 2 can be used to solve the problem.

6 0

3 years ago

Read 2 more answers

Pls pls Pls pls Pls Pls help me pls pls pls pls pls pls ASAP pls pls pls Pls pls Pls pls Pls Pls help me pls pls I WILL GIVE BRA

MAVERICK [17]

Answer:

y = -3x+8

Step-by-step explanation:

formula = y=mx+b

slope is m

y-intercept is b

4 0

3 years ago

Three numbers in the interval $\left[0,1\right]$ are chosen independently and at random. What is the probability that the chosen

k0ka [10]

Let $a,b,c$ be the randomly selected lengths. Without loss of generality, suppose $a[tex]P(A + B \ge C) = P(A + B - C \ge 0)$

where $A,B,C$ are independent random variables with the same uniform distribution on [0, 1].

By their mutual independence, we have

$P(A=a,B=b,C=c) = P(A=a) \times P(B=b) \times P(C=c)$

so that the joint density function is

$P(A=a,B=b,C=c) = \begin{cases}1 & \text{if }(a,b,c)\in[0,1]^3 \\ 0 & \text{otherwise}\end{cases}$

where $[0,1]^3=[0,1]\times[0,1]\times[0,1]$ is the cube with vertices at (0, 0, 0) and (1, 1, 1).

Consider the plane

$a + b - c = 0$

with $(a,b,c)\in\Bbb R^3$ . This plane passes through (0, 0, 0), (1, 0, 1), and (0, 1, 1), and thus splits up the cube into one tetrahedral region above the plane and the rest of the cube under it. (see attached plot)

The point (0, 0, 1) (the vertex of the cube above the plane) does not belong the region $a+b-c\ge0$ , since $0+0-1=-1$ . So the probability we want is the volume of the bottom "half" of the cube. We could integrate the joint density over this set, but integrating over the complement is simpler since it's a tetrahedron.

Then we have

$\displaystyle P(A+B-C\ge0) = 1 - P(A+B-C < 0) \\\\ ~~~~~~~~ = 1 - \int_0^1\int_0^{1-a}\int_{a+b}^1 P(A=a,B=b,C=c) \, dc\,db\,da \\\\ ~~~~~~~~ = 1 - \int_0^1 \int_0^{1-a} (1 - a - b) \, db \, da \\\\ ~~~~~~~~ = 1 - \int_0^1 \frac{(1-a)^2}2\,da \\\\ ~~~~~~~~ = 1 - \frac16 = \boxed{\frac56}$

5 0

1 year ago

John had a roll containing 24 2/3 feet of wrapping paper. He wants to divide it into 11 pieces. First, though, he must cut off 5

tatyana61 [14]

Given

John had a roll containing of wrapping paper

$=24\tfrac{2}{3}$

First, he must cut off 5/6 foot because it was torn.

He wants to divide it into 11 pieces.

To proof

As given in question

John had a roll containing of wrapping paper

$=24\tfrac{2}{3}$

$=\frac{74}{3}$ ( solving the mixed fraction form)

First, though, he must cut off 5/6 foot

than we get

$= \frac{74}{3} - \frac{5}{6}$

Taking the L.C.M of 3 and 6 is 6

$=\frac{148-5}{6}$

$=\frac{143}{6}$

=23.83 feet (approx)

He wants to divide it into 11 piece

length of the each piece

$=\frac{23.83}{11}$

length of the each pieces be 2.16 feet ( approx)

Hence proved

6 0

3 years ago

Read 2 more answers

what are the measure measurements of angels a b and c show your work and explain your work and explain your answers

julia-pushkina [17]

A = 40° (Vertically opposite angles)

b = 180 - 90 - 40 = 50° (Angles in a triangle add up to 180)

c = 180 - 65 = 115° (Adjacent angles on a straight line)

5 0

3 years ago

-5(x + 1) s 30. What is a possible value of x? help me pls