How to Simplify (2v)^4

You are given the equation \sqrt{76+n } and \sqrt{2n + 26}. What is the smallest value of n that will make each number rational?

Diano4ka-milaya [45]

Answer:

A proof that square root of 2 is irrational. ... An equation x² = a, and the principal square root ... The number under the radical sign is called the radicand. ... the given square numbers, each product of square numbers is equal to what square ... are relatively prime -- and it will be impossible to divide n· n into m· m and get 2.

Step-by-step explanation:

4 0

2 years ago

I can’t solve this. Please explain!

babunello [35]

Sorry I can’t help....

6 0

3 years ago

Read 2 more answers

Which equation represents a line which is perpendicular to the y-axis?

erma4kov [3.2K]

Answer:

y = 5

Step-by-step explanation:

The y-axis and any lines that are parallel to it are vertical. Therefore, a line that would be perpendicular to the y-axis would need to be horizontal. All horizontal lines have a slope equal to 0 and are represented by the equation y = a constant, or a single number with no variables.

The only option that follows that format is y = 5, thus that is the answer.

6 0

2 years ago

In the xy-coordinate plane shown, line L passes through point C and has a slope of -2. A dilation of line {with center A and a s

masya89 [10]

Answer:

C' (-4 , 2)

slope of new line = line "L" = -2

Step-by-step explanation:

C (x₀,y₀): (0,2) dilation center (h,k): (2,2) scale factor (sf): 3

C' (x,y)

x = 2 + 3*(0-2) = -4 x = h + sf*(x₀ - h)

y = 2 + 3*(2-2) = 2 y = k + sf*(y₀ - k)

C' (-4 , 2)

slope of new line = line "L" = -2

8 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

2 years ago