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3 years ago

Solve the x 1/3x + 5> 10

Mathematics

1 answer:

kotykmax [81]3 years ago

5 0

Answer:

The solution to the inequality is:

$\frac{1}{3}x+5>10\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x>15\:\\ \:\mathrm{Interval\:Notation:}&\:\left(15,\:\infty \:\right)\end{bmatrix}$

The solution graph is also attached below.

Step-by-step explanation:

Given

We are given the expression

$\frac{1}{3}x+5>10$

To determine

Solve for x

Given the expression

$\frac{1}{3}x+5>10$

Subtract 5 from both sides

$\frac{1}{3}x+5-5>10-5$

Simplify

$\frac{1}{3}x>5$

Multiply both sides by 3

$3\cdot \frac{1}{3}x>5\cdot \:3$

Simplify

$x>15$

Thus, we conclude that:

$x>15$

Therefore, the solution to the inequality is:

$\frac{1}{3}x+5>10\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x>15\:\\ \:\mathrm{Interval\:Notation:}&\:\left(15,\:\infty \:\right)\end{bmatrix}$

The solution graph is also attached below.

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Assume that when adults with smartphones are randomly selected, 48% use them in meetings or classes. If 8 adult smartphone use

Lisa [10]

Answer:

The probability is 2.2%

Step-by-step explanation:

The probability that a randomly selected adult uses the telephone in classes is always equal to 0.48.

If x represents the number of adults selected from among the 8 that the telephone uses in classes, then x is a discrete random variable that follows a binomial distribution, with:

p = 0.48

q = 0.52

n = 8

We want to find the probability of x = 5

$P(x) = \frac{n!}{x!(n-x)!} *p ^ n *q ^{n-x}\\\\ P(x = 5) = \frac{8!}{5!(8-5)!} * 0.48 ^ 8 * 0.52 ^ 3\\\\ P(x = 5) = 0.02218$

The probability is 2.2%

6 0

3 years ago

Use the equation of the circle centered at the origin and

Serggg [28]

Answer:

Step-by-step explanation:

The standard form of a circle is

$(x-h)^2+(y-k)^2=r^2$

If we are given x and y as 0 and 5 respectively, and we are also told that the center is (0, 0), our h and k are both 0. Filling in x, y, h, and k we can find the radius. So let's do that:

$(0-0)^2+(5-0)^2=r^2$ and

$0^2+5^2=r^2$ so

$r^2=25$ Our circle's equation is

$x^2+y^2=25$

Since the point in question, (4, 4), lies in the first quadrant, we will concentrate on that quadrant only. To fall within the circle, we can set up an inequality and test the point (4, 4). If it lies ON the circle then the equality would be true. Let's try that first:

$4^2+4^2=25$

Obviously, 16 + 16 does not equal 25, so that point (4, 4) does not lie ON the circle. In fact from that statement alone, we can determine that the point lies outside the circle because

$4^2+4^2>25$