Simplify, write without absolute value sign<br><br>

The probability that a can of paint contains contamination is 3.23%, and the probability of a mixing error is 2.4%. The probabil

stich3 [128]

Answer:

4.6%.

Step-by-step explanation:

The probability that a can of paint contains contamination(C) is 3.23%

P(C)=3.23%

The probability of a mixing(M) error is 2.4%.

P(M)=2.4%

The probability of both is 1.03%.

$P(C \cap M)=1.03\%$

We want to determine the probability that a randomly selected can has contamination or a mixing error. i.e. $P(C \cup M)$

In probability theory:

$P(C \cup M) = P(C)+P(M)-P(C \cap M)\\P(C \cup M)=3.23+2.4-1.03\\P(C \cup M)=4.6\%$

The probability that a randomly selected can has contamination or a mixing error is 4.6%.

3 0

3 years ago

24. Find the area of the shaded region:

Monica [59]

Answer:

A = 5x³ + 17x² - 35x - 30

Step-by-step explanation:

the shaded area (A) is calculated as

A = outer area - inner area

= (5x + 3)(x² + 4x - 10) - 3x(2x - 1) ← distribute parenthesis

= 5x³ + 20x² - 50x + 3x² + 12x - 30 - (6x² - 3x)

= 5x³ + 23x² - 38x - 30 - 6x² + 3x ← collect like terms

= 5x³ + 17x² - 35x - 30

7 0

2 years ago

Read 2 more answers

F(x)=3x-1 g(x) x^3 -5x^2 <br> Find (f o g)(x)

Rudik [331]

plz plz mark brainliest

7 0

3 years ago

Read 2 more answers

The hyperbolic orbit of a comet is represented on a coordinate plane with center at (0, 4). One branch has a vertex at (0, 10) a

Anuta_ua [19.1K]

Remember that

If the given coordinates of the vertices and foci have the form (0,10) and (0,14)

then

the transverse axis is the y-axis

so

the equation is of the form

(y-k)^2/a^2-(x-h)^2/b^2=1

In this problem

center (h,k) is equal to (0,4)

(0,a-k)) is equal to (0,10)

a=10-4=6

(0,c-k) is equal to (0,14)

c=14-4=10

Find out the value of b

b^2=c^2-a^2

b^2=10^2-6^2

b^2=64

therefore

the equation is equal to

<h2>(y-4)^2/36-x^2/64=1</h2><h2>the answer is option A</h2>

6 0

1 year ago

Which is the interval notation represent the domain?

Kazeer [188]

Answer:

The first choice, $(-\infty, 7) \cup (7, \infty)$ .

Step-by-step explanation:

The dashed vertical line is a vertical asymptote. The x-coordinate of all points on this vertical asymptote are equal to 7. In other words, when the x-value of a point on the graph approaches $7$ , its y-value approaches infinity (or negative infinity.)

Either way, the graph is not defined for $x = 7$ . The point $7$ should thus be excluded from the domain of the graph.

The graph is apparently defined for all other x-values. The domain of the function should thus be all real numbers with the exception of $x = 7$ . Here's how to write that in interval notation:

The set of all real numbers can be expressed as the interval $(-\infty, \infty)$ . Note that infinity $\infty$ (or $(- \infty)$ itself isn't a real number. The round brackets indicate that both endpoints are excluded from the from the interval.
The set of all real numbers less than (not equal to) $7$ is $(-\infty,7 )$ (both ends are excluded.) The set of all real numbers greater than (not equal to) $7$ is $(7, \infty)$ .
The set of all real numbers that is not equal to $7$ is the union of all real numbers less than $7$ and all real numbers greater than $7$ . The union operator $\cup$ connects two intervals.

In other words, the domain is $(-\infty, 7) \cup (7, \infty)$ .

5 0

3 years ago

Simplify, write without absolute value sign | x+3 |, if x > 5