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3 years ago

Employers may use your credit report to decide wether to hire you. True or False

Mathematics

1 answer:

g100num [7]3 years ago

5 0

Answer:

False

Step-by-step explanation:

the reason why is because why would your employers have any right, or have any reason to go into your credit report to hire you?

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Do anyoneeeeeee know this??????????????????????????????????????

Dennis_Churaev [7]

Answer is 12

K=y/x (rise over run)

The same rise over run happens to the next point, up 12 over 1

5 0

2 years ago

Read 2 more answers

2. To estimate the mean age for a population of 4000 employees, a simple random sample of 40 employees is selected. a. Would you

ANEK [815]

Answer:

Step-by-step explanation:

Size of the population, N = 4000

Size of the sample, n = 40

n/N = 40/4000 = 0.01

0.01 is less than 0.05 and hence we would not us the finite population correction factor in calculating the standard error of the mean.

Population standard deviation is σ = 8.2

So, the standard error of x’ using the finite population correction factor is give by

σ(x’) = √[(N - n)/(N - 1)] x (sigma/√n)

σ(x’) = √[(4000 - 40)/(4000 - 1)] x (8.2/√40)

σ(x’) = 1.29

Standard error of x’ without using the finite population correction factor is

σ(x’) = σ/√n

σ(x’) = 8.2/√40 = 1.2965

There is little difference between the two values of the standard error. So we can ignore the population correction factor.

Let the population mean be μ

Probability that the sample mean will be within =-2 of the population mean is

P(μ– 2 < x’ < μ + 2)

At x’ = μ – 2 , we have

z = (μ – 2 – μ)/1.2965

z = -1.54

at x’ = μ + 2, we have

z = (μ + 2 – μ)/1.2965

z = 1.54

So the required probability is

P(μ – 2 < x’ < μ + 2) = p(-1.54< z < 1.54)

P(μ – 2 < x’ < μ + 2) = p(z < 1.54) – p(z < -1.54)

P(μ – 2 < x’ < μ + 2) = 0.9382 – 0.0618

P(μ – 2 < x’ < μ + 2) = 0.8764

7 0

3 years ago

Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x' (t) is its velocity, and x'

vodka [1.7K]

Answer:

a) $v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9$

$a(t) = x''(t) = v'(t) =6t-12$

b) $0$

c) $a(t) = x''(t) = v'(t) =6t-12$

When the acceleration is 0 we have:

$6t-12=0, t =2$

And if we replace t=2 in the velocity function we got:

$v(t) = 3(2)^2 -12(2) +9=-3$

Step-by-step explanation:

For this case we have defined the following function for the position of the particle:

$x(t) = t^3 -6t^2 +9t -5 , 0\leq t\leq 10$

Part a

From definition we know that the velocity is the first derivate of the position respect to time and the accelerations is the second derivate of the position respect the time so we have this:

$v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9$

$a(t) = x''(t) = v'(t) =6t-12$

Part b

For this case we need to analyze the velocity function and where is increasing. The velocity function is given by:

$v(t) = 3t^2 -12t +9$

We can factorize this function as $v(t)= 3 (t^2- 4t +3)=3(t-3)(t-1)$

So from this we can see that we have two values where the function is equal to 0, t=3 and t=1, since our original interval is $0\leq t\leq 10$ we need to analyze the following intervals:

$0< t$

For this case if we select two values let's say 0.25 and 0.5 we see that

$v(0.25) =6.1875, v(0.5)=3.75$

And we see that for $a=0.5 >0.25=b$ we have that f(b)>f(a) so then the function is decreasing on this case.

$1$

We have a minimum at t=2 since at this value w ehave the vertex of the parabola :

$v_x =-\frac{b}{2a}= -\frac{-12}{2*3}= -2$

And at t=-2 v(2) = -3 that represent the minimum for this function, we see that if we select two values let's say 1.5 and 1.75

v(1.75) =-2.8125< -2.25= v(1.5) so then the function sis decreasing on the interval 1<t<2

$2$

We see that the function would be increasing.

$3$

For this interval we will see that for any two points a,b with $a>b$ we have $f(a)>f(b)$ for example let's say a=3 and b =4

$f(a=3) =0 , f(b=4) =9 , f(b)>f(a)$

The particle is moving to the right then the velocity is positive so then the answer for this case is: $0$

Part c

$a(t) = x''(t) = v'(t) =6t-12$

When the acceleration is 0 we have:

$6t-12=0, t =2$

And if we replace t=2 in the velocity function we got:

$v(t) = 3(2)^2 -12(2) +9=-3$

5 0

3 years ago

In order to test for the significance of a regression model involving 5 independent variables and 123 observations, the numerato

allochka39001 [22]

Answer: The numerator and denominator degrees of freedom (respectively) for the critical value of F are 4 and 118 .

Step-by-step explanation:

We know that , for critical value of F, degrees of freedom for numerator = k-1

and for denominator = n-k, where n= Total observations and k = number of independent variables.

Here, Numbers of independent variables(k) = 5

Total observations (n)= 123

So, Degrees of freedom for numerator = 5-1=4

Degrees of freedom for denominator =123-5= 118

Hence, the numerator and denominator degrees of freedom (respectively) for the critical value of F are 4 and 118 .

5 0

3 years ago

Which equation represents the parabola shown on the

EastWind [94]

Answer:

y2=-8x is what's graphed so B.

Step-by-step explanation:

(I think you mistyped the choice)

3 0

3 years ago

Read 2 more answers

Employers may use your credit report to decide wether to hire you. True or False​

Employers may use your credit report to decide wether to hire you. True or False