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2 months ago

12-5x-4kx=y solve for x

Mathematics

1 answer:

zzz [12.3K]2 months ago

3 0

Hello!

$12 - 5x - 4kx = y

-4kx - 5x = y - 12 

Factor out variable x 

x(-4k-5)=y-12

x = (y+12)/(4k+5)$

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Which distribution exhibits a negative skew?

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In graphs showcasing investment returns displaying positive skewness, this indicates that: mean > median > mode. Conversely, a negative skewness reveals the relationship: mean < median < mode.

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4 0

16 days ago

Read 2 more answers

Find the distance from (4, −7, 6) to each of the following.

Zina [12379]

Answer:

(a) 6 units

(b) 4 units

(d) 9.22 units

(e) 7.21 units

(f) 8.06 units

Step-by-step explanation:

The distance between two points, (x₁, y₁, z₁) and (x₂, y₂, z₂), can be calculated using;

d = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²]

According to the problem;

(a) The distance from (4, -7, 6) to the xy-plane

The xy-plane corresponds to where z equals 0, so

xy-plane = (4, -7, 0).

Thus, the distance d is calculated from (4, -7, 6) to (4, -7, 0)

d = √[(4 - 4)² + (-7 - (-7))² + (0 - 6)²]

d = √[(0)² + (0)² + (-6)²]

d = √(-6)²

d = √36

d = 6

Thus, the distance to the xy-plane is 6 units

(b) The distance from (4, -7, 6) to the yz-plane

The yz-plane is located where x is 0, hence

yz-plane = (0, -7, 6).

So, the distance d is from (4, -7, 6) to (0, -7, 6)

d = √[(4 - 0)² + (-7 - (-7))² + (6 - 6)²]

d = √[(4)² + (0)² + (0)²]

d = √(4)²

d = √16

d = 4

Thus, the distance to the yz-plane is 4 units

(c) The distance from (4, -7, 6) to the xz-plane

The xz-plane exists where y is 0, meaning

xz-plane = (4, 0, 6).

The distance d from (4, -7, 6) to (4, 0, 6)

d = √[(4 - 4)² + (-7 - 0)² + (6 - 6)²]

d = √[(0)² + (-7)² + (0)²]

d = √[(-7)²]

d = √49

d = 7

Thus, the distance to the xz-plane is 7 units

(d) The distance from (4, -7, 6) to the x-axis

The x-axis is defined by y and z being 0, which implies

x-axis = (4, 0, 0).

Thus, the distance d is from (4, -7, 6) to (4, 0, 0)

d = √[(4 - 4)² + (-7 - 0)² + (6 - 0)²]

d = √[(0)² + (-7)² + (6)²]

d = √[(-7)² + (6)²]

d = √[(49 + 36)]

d = √(85)

d = 9.22

Hence, the distance to the x-axis is 9.22 units

(e) The distance from (4, -7, 6) to the y-axis

The y-axis is defined where x and z are both 0, thus

y-axis = (0, -7, 0).

Thus, the distance d is from (4, -7, 6) to (0, -7, 0)

d = √[(4 - 0)² + (-7 - (-7))² + (6 - 0)²]

d = √[(4)² + (0)² + (6)²]

d = √[(4)² + (6)²]

d = √[(16 + 36)]

d = √(52)

d = 7.22

Thus, the distance to the y-axis is 7.21 units

(f) The distance from (4, -7, 6) to the z-axis

The z-axis is defined by x and y being 0, which gives

z-axis = (0, 0, 6).

Thus, the distance d is calculated from (4, -7, 6) to (0, 0, 6)

d = √[(4 - 0)² + (-7 - 0)² + (6 - 6)²]

d = √[(4)² + (-7)² + (0)²]

d = √[(4)² + (-7)²]

d = √[(16 + 49)]

d = √(65)

d = 8.06

Thus, the distance to the z-axis is 8.06 units

5 0

1 month ago

Which expressions are equivalent to RootIndex 3 StartRoot 128 EndRoot Superscript x? Select three correct answers.

AnnZ [12381]

Answer:

- 128 Superscript StartFraction 3 Over x EndFraction

- (4RootIndex 3 StartRoot 2 EndRoot)x

- (4 (2 Superscript one-third Baseline) ) Superscript x

Step-by-step explanation:

Considerando la ecuación dada $(\sqrt[3]{128} )^{x}\\$

De acuerdo con una de las leyes de índices,

$(\sqrt[a]{m} )^{b}\\= (\sqrt{m})^\frac{b}{a}$

Aplicando esta ley a la pregunta;

$(\sqrt[3]{128} )^{x}\\ = {128} ^\frac{x}{3}\\ \\= (\sqrt[3]{64*2})^{x} \\ = (4\sqrt[3]{2})^{x} \\= (4(2^{1/3} )^{x} )$

<pLos siguientes son ciertos de acuerdo con el cálculo presentado

128 Superscript StartFraction 3 Over x EndFraction

(4RootIndex 3 StartRoot 2 EndRoot)x

(4 (2 Superscript one-third Baseline) ) Superscript x

5 0

1 month ago

Read 2 more answers

Use the normal approximation to the binomial distribution to answer this question. Fifteen percent of all students at a large un

AnnZ [12381]

Answer: 0.1289

Step-by-step explanation:

Given: The proportion of students absent on Mondays at a large university.: $p=0.15$

Sample size: $n=12$

Mean: $\mu=np=12\times0.15=1.8$

Standard deviation = $\sigma=\sqrt{np(1-p)}$

$\Rightarrow\ \sigma=\sqrt{12(0.15)(1-0.15)}=1.23693168769\approx1.2369$

Let x represent a binomial variable.

Referencing the standard normal distribution table,

$P(x=4)=P(x\leq4)-P(x\leq3)$ (1)

Z score for normal distribution:-

$z=\dfrac{x-\mu}{\sigma}$

For x=4

$z=\dfrac{4-1.8}{1.2369}\approx1.78$

For x=3

$z=\dfrac{3-1.8}{1.2369}\approx0.97$

Thus, from (1)

$P(x=4)=P(z\leq1.78)-P(z\leq0.97)\\\\=0.962462-0.8339768\approx0.1289$

Consequently, the likelihood of four students being absent = $0.1289$

3 0

1 month ago

Axline Computers manufactures personal computers at two plants, one in Texas and the other in Hawaii. The Texas plant has 40 emp

Zina [12379]

Answer:

a) The likelihood that none of the sampled employees are from the Hawaii plant is 1.74%.

b) The chance that exactly 1 employee from the sample is found working in the Hawaii plant is 8.70%.

c) There is an 89.56% chance that 2 or more employees in the sample are from the Hawaii plant.

d) The probability that 9 employees from the sample are working at the Texas plant is 8.70%.

Step-by-step explanation:

Each employee has two potential employment locations: either Texas or Hawaii. Thus, the binomial probability distribution can be utilized to solve this scenario.

Binomial probability distribution

This distribution defines the probability of achieving exactly x successes in n trials where there are only two outcomes.

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

Here, $C_{n,x}$ denotes the number of ways to choose x objects from a set of n, represented by the subsequent formula.

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

And p is the probability of success occurring.

In this context, we know:

The sample comprises 10 employees, therefore $n = 10$ .

a. Calculate the probability that none of the sampled employees are from the Hawaii plant (to 4 decimals)?

Given that 20 out of 60 employees are based in Hawaii:

$p = \frac{20}{60} = 0.333$

We aim to find P(X = 0).

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

$P(X = 0) = C_{10,0}.(0.333)^{0}.(0.667)^{10} = 0.0174$

Thus, the likelihood that none in the sample are from Hawaii stands at 1.74%.

b. Calculate the probability that 1 employee from the sample is from the Hawaii plant?

This is represented as P(X = 1).

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

$P(X = 1) = C_{10,1}.(0.333)^{1}.(0.667)^{9} = 0.0870$

Therefore, there is an 8.70% possibility that 1 employee in the sample comes from Hawaii.

c. Calculate the probability that 2 or more employees in the sample are from the Hawaii plant?

We can observe two scenarios: either fewer than 2 employees are from Hawaii or 2 and beyond. The combined probabilities equal decimal 1. So:

$P(X < 2) + P(X \geq 2) = 1$

We seek to find $P(X \geq 2)$ .

$P(X \geq 2) = 1 - P(X < 2)$

From problems a and b, we possess values for both probabilities.

$P(X < 2) = P(X = 0) + P(X = 1) = 0.0174 + 0.0870 = 0.1044$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.1044 = 0.8956$

Accordingly, the chance that 2 or more employees in this sample operate at the Hawaii plant is 89.56%.

d. Calculate the likelihood that 9 employees in the sample are working at the Texas plant?

This corresponds to the probability found in part b for 1 employee working in Hawaii.

Consequently, there is an 8.70% chance that 9 employees belong to the Texas plant.

6 0

1 month ago