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3 years ago

Select the exclusion that fits best with this problem. 13x^6/51x^4

Mathematics

1 answer:

Damm [24]3 years ago

5 0

ANSWER

$x \ne0$

EXPLANATION

The given problem is

$\frac{13 {x}^{6} }{ 51 {x}^{4} }$

This is defined if and only if

$x \ne0$

Even the given expression can be simplified to obtain:

$\frac{13 {x}^{2} }{ 51}$

The exclusion is

$x \ne0$

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Find the slope of the line passing through these points: (5,2) and (-10,12)

Marat540 [252]

Answer:

-2/3

Step-by-step explanation:

The slope of a line is found by

m= (y2-y1)/(x2-x1)

= (12-2)/(-10 -5)

= 10/-15

= -2/3

3 0

3 years ago

Use all 4 operations and at least one exponent to write and expression that has a value of 10

Yuki888 [10]

5+5=10

15-5=10

5x2=10

20/2=10

10^1 (10 to the power of 1) or 0.1^-1 (0.1 to the power of -1)

7 0

2 years ago

A skateboarding ramp is 13in. high and rises at an angle of 13°. How long is the base of the ramp? Round to the nearest inch.

charle [14.2K]

<h3>Answer: Approximately 56 inches</h3>

===========================================================

Explanation:

The diagram is shown below. We know the vertical part of the triangle is 13 inches, which is the side opposite the reference angle 13 degrees. The adjacent side is unknown. We'll call it x. This is how long the base of the ramp is, which is the horizontal distance along the entire ramp. This distance is on the ground. The ramp itself is the hypotenuse but it seems like your teacher isn't wanting to know this value. So we'll ignore the hypotenuse.

We'll use the tangent rule to connect the opposite and adjacent sides.

tan(angle) = opposite/adjacent

tan(13) = 13/x

x*tan(13) = 13

x = 13/tan(13)

x = 56.309186365694

x = 56 inches is the approximate horizontal distance underneath the ramp

5 0

2 years ago

This exercise illustrates that poor quality can affect schedules and costs. A manufacturing process has 90 customer orders to fi

svp [43]

Answer:

a) 0.0645 = 6.45% probability that the 90 orders can be filled without reordering components.

b) 0.4062 = 40.62% probability that the 100 orders can be filled without reordering components.

c) 0.9034 = 90.34% probability that the 100 orders can be filled without reordering components

Step-by-step explanation:

For each component, there are only two possible outcomes. Either it is defective, or it is not. The components can be assumed to be independent. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

3% of the components are identified as defective

This means that $p = 0.03$

a. If the manufacturer stocks 90 components, what is the probability that the 90 orders can be filled without reordering components?

0 defective in a set of 90, which is $P(X = 0)$ when $n = 90$ . So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{90,0}.(0.03)^{0}.(0.97)^{90} = 0.0645$

0.0645 = 6.45% probability that the 90 orders can be filled without reordering components.

b. If the manufacturer stocks 102 components, what is the probability that the 100 orders can be filled without reordering components?

At most 102 - 100 = 2 defective in a set of 102, so $P(X \leq 2)$ when $n = 102$

Then

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{102,0}.(0.03)^{0}.(0.97)^{102} = 0.0447$

$P(X = 1) = C_{102,0}.(0.03)^{1}.(0.97)^{101} = 0.1411$

$P(X = 2) = C_{102,2}.(0.03)^{2}.(0.97)^{100} = 0.2204$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0447 + 0.1411 + 0.2204 = 0.4062$

0.4062 = 40.62% probability that the 100 orders can be filled without reordering components.

c. If the manufacturer stocks 105 components, what is the probability that the 100 orders can be filled without reordering components?

At most 105 - 100 = 5 defective in a set of 105, so $P(X \leq 5)$ when $n = 105$

Then

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{105,0}.(0.03)^{0}.(0.97)^{105} = 0.0408$

$P(X = 1) = C_{105,0}.(0.03)^{1}.(0.97)^{104} = 0.1326$

$P(X = 2) = C_{105,2}.(0.03)^{2}.(0.97)^{103} = 0.2133$

$P(X = 3) = C_{105,3}.(0.03)^{3}.(0.97)^{102} = 0.2265$

$P(X = 4) = C_{105,4}.(0.03)^{4}.(0.97)^{101} = 0.1786$

$P(X = 5) = C_{105,5}.(0.03)^{5}.(0.97)^{100} = 0.1116$

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0408 + 0.1326 + 0.2133 + 0.2265 + 0.1786 + 0.1116 = 0.9034$

0.9034 = 90.34% probability that the 100 orders can be filled without reordering components

3 0

3 years ago

I have to find slope-intercept equation

adoni [48]

Answer:

y = 4x - 9

Step-by-step explanation:

y = mx + b

m = slope

b = y-intercept

the y-intercept is when the line crosses the y axis, and in this case we can see it at "-9"

The slope we can do rise over run "r/r" ot "y/x"

r/r = 4/1 = 4

The line is positive since its going up

y = 4x - 9

Hope this helped!

Have a supercalifragilisticexpialidocious day!

7 0

3 years ago