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3 years ago

Write a formula for y in terms of x

Mathematics

1 answer:

garik1379 [7]3 years ago

4 0

Answer:

question is incomplete

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The weekly amount spent by a small company for in-state travel has approximately a normal distribution with mean $1450 and stand

Llana [10]

Answer:

0.0903

Step-by-step explanation:

Given that :

The mean = 1450

The standard deviation = 220

sample mean = 1560

$P(X > 1560) = P( Z > \dfrac{x - \mu}{\sigma})$

$P(X > 1560) = P(Z > \dfrac{1560 - 1450}{220})$

$P(X > 1560) = P(Z > \dfrac{110}{220})$

P(X> 1560) = P(Z > 0.5)

P(X> 1560) = 1 - P(Z < 0.5)

From the z tables;

P(X> 1560) = 1 - 0.6915

P(X> 1560) = 0.3085

Let consider the given number of weeks = 52

Mean $\mu_x$ = np = 52 × 0.3085 = 16.042

The standard deviation = $\sqrt {n \time p (1-p)}$

The standard deviation = $\sqrt {52 \times 0.3085 (1-0.3085)}$

The standard deviation = 3.3306

Let Y be a random variable that proceeds in a binomial distribution, which denotes the number of weeks in a year that exceeds $1560.

Then;

Pr ( Y > 20) = P( z > 20)

$Pr ( Y > 20) = P(Z > \dfrac{20.5 - 16.042}{3.3306})$

$Pr ( Y > 20) = P(Z >1 .338)$

From z tables

P(Y > 20) $\simeq$ 0.0903

7 0

3 years ago

Equivalent of -3(5+x)+4x

Natasha_Volkova [10]

Answer:

-15 + x

Step-by-step explanation:

5 0

3 years ago

Ik this is rlly easy im just to lazy to do it

True [87]

The answer 12/72 because you multiply

4 0

3 years ago

ASAP Please answer. Three students sit in the first row. In the second row, 3 students sit behind each of those students, and so

murzikaleks [220]

Answer:

3^2, 9 students.

Step-by-step explanation:

I'm not 100 percent sure about this but I'm pretty sure that because there are 3 rows you can just do 3^2 which is 9.

3 0

3 years ago

Read 2 more answers

Please help me with this

Ne4ueva [31]

Answer:

Option c

Step-by-step explanation:

A system of equations are given to us. And we need to solve them . The given system is

$\begin{cases} y = 2x - 3\dots 1 \\ y = x^2 - 3\dots 2 \end{cases}$

We numbered the equations here . Now put the value of equation 1 in equation 2 that is substituting y = 2x - 3 in eq. 2 .

$\implies y = x^2 - 3 \\\\\implies 2x - 3 = x^2 - 3 \\\\\implies x^2 - 2x = 0 \\\\\implies x(x-2) = 0 \\\\\implies\red{ x = 0 , 2 }$

We got two values of x as 0 & 2 . Alternatively substituting these values we have ,

$\implies y = 2 x - 3 \\\\\implies y = 2(0)-3 \qquad or \qquad y = 2(2)-3 \\\\\implies y = 0-3 \qquad or \qquad 4 - 3 \\\\\implies \red{ y = -3 , 1 }$

Thefore the required answer is ,

$\red{Option\:c} \begin{cases} (0,-3) \\ (2,1) \end{cases}$

4 0

3 years ago

Read 2 more answers