All categories

3 years ago

What is the solution to the linear equation? 2/3x-1/2=1/3+5/6x

2 answers:

7 0

2/3x - 1/2 = 1/3 + 5/6x....multiply the entire equation by the common denominator of 6...this will get rid of the fractions.

6(2/3x - 1/2) = 6(1/3 + 5/6x) =
4x - 3 = 2 + 5x...subtract 5x from both sides
4x - 5x - 3 = 2...add 3 to both sides
4x - 5x = 2 + 3...combine terms
-x = 5...multiply by -1 to make x positive
x = -5

strojnjashka [21]3 years ago

3 0

Answer:

x=-5

Step-by-step explanation:

You might be interested in

What is the simplified answer to this problem 12 > -3x + 6

Verdich [7]

In this question, you're simplifying the inequality by solving for x.

Solve for x:

12 > -3x + 6

<em>flip the equation:</em>

-3x + 6 < 12

<em>subtract 6 from both sides</em>

-3x < 6

<em>divide both sides by -3, while also flipping the inequality</em>

x > -2

Answer:

x > -2

8 0

3 years ago

A store that sells shoes buys them from a manufacturer at a wholesale price of $57. The store's markup rate is 50%. What price d

kogti [31]

The answer is $85.50

57 x 50% or 57 x .5 = 28.50

Add $28.50 to $57 and you get $85.50

4 0

3 years ago

Find two complex numbers that have a sum of i10 a different of -4and a product of -29

lianna [129]

Step-by-step explanation:

Let the complex numbers be $a+ib\textrm{ and }c+id$ .

Given, sum is $10i$ , difference is $-4$ and product is $-29$ .

$(a+c)+i(b+d)=10i$ ⇒ $a+c=0,b+d=10$

$(a-c)+i(b-d)=-4$ ⇒ $a-c=-4,b-d=0$

$a=-2,c=2,b=5,d=5$

$(a+ib)(c+id)=(-2+5i)(2+5i)=-4-25=-29$

Hence, all three equations are consistent yielding the complex numbers $-2+5i\textrm{ and }2+5i$ .

8 0

3 years ago

Please guys i really need your help and please explain how you got it please

lina2011 [118]

The answer would be C because in order to find the area of a figure, you would need to multiply its height by its width. finding the area of a figure is the same as finding how many units are in a figure. because this is a square and all the sides are equal, by counting the total number of units, you can find the area.

5 0

3 years ago

An instructor who taught two sections of engineering statistics last term, the first with 25 students and the second with 35, de

Amiraneli [1.4K]

Answer:

a) P=0.1721

b) P=0.3528

c) P=0.3981

Step-by-step explanation:

This sampling can be modeled by a binominal distribution where p is the probability of a project to belong to the first section and q the probability of belonging to the second section.

a) In this case we have a sample size of n=15.

The value of p is p=25/(25+35)=0.4167 and q=1-0.4167=0.5833.

The probability of having exactly 10 projects for the second section is equal to having exactly 5 projects of the first section.

This probability can be calculated as:

$P=\frac{n!}{(n-k)!k!}p^kq^{n-k}= \frac{15!}{(10)!5!}\cdot 0.4167^5\cdot0.5833^{10}=0.1721$

b) To have at least 10 projects from the 2nd section, means we have at most 5 projects for the first section. In this case, we have to calculate the probability for k=0 (every project belongs to the 2nd section), k=1, k=2, k=3, k=4 and k=5.

We apply the same formula but as a sum:

$P(k\leq5)=\sum_{k=0}^{5}\frac{n!}{(n-k)!k!}p^kq^{n-k}$

Then we have:

$P(k=0)=0.0003\\P(k=1)=0.0033\\P(k=2)=0.0165\\P(k=3)=0.0511\\P(k=4)=0.1095\\P(k=5)=0.1721\\\\P(k\leq5)=0.0003+0.0033+0.0165+0.0511+0.1095+0.1721=0.3528$

c) In this case, we have the sum of the probability that k is equal or less than 5, and the probability tha k is 10 or more (10 or more projects belonging to the 1st section).

The first (k less or equal to 5) is already calculated.

We have to calculate for k equal to 10 or more.

$P(k\geq10)=\sum_{k=10}^{15}\frac{n!}{(n-k)!k!}p^kq^{n-k}$

Then we have

$P(k=10)=0.0320\\P(k=11)=0.0104\\P(k=12)=0.0025\\P(k=13)=0.0004\\P(k=14)=0.0000\\P(k=15)=0.0000\\\\P(k\geq10)=0.032+0.0104+0.0025+0.0004+0+0=0.0453$

The sum of the probabilities is

$P(k\leq5)+P(k\geq10)=0.3528+0.0453=0.3981$

8 0

3 years ago