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3 years ago

Please answer these for me <3

Mathematics

1 answer:

gtnhenbr [62]3 years ago

6 0

8-21

9- -16 i think lemme do the next ones

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What is the distance between -6 and 2 on a number line

Likurg_2 [28]

I made a number line and drew it out -6,-5,-4,-3,-2,-1,0,1,2 and i counted how many i had to move over to get to 2 from -6 and i got 8

7 0

3 years ago

HELP PLS<br><br> The function of F is defined by f(x)=x+2<br> find F(n-4)

viktelen [127]

The function of F is defined by f(x) = x + 2, the value of f(n - 4) is given as f(n - 4) = n - 4 + 2 = n - 2

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

Given the equation:

f(x) = x + 2

The value of f(n - 4) is given as:

f(n - 4) = n - 4 + 2 = n - 2

The function of F is defined by f(x) = x + 2, the value of f(n - 4) is given as f(n - 4) = n - 4 + 2 = n - 2

Find out more on equation at: brainly.com/question/2972832

#SPJ1

6 0

2 years ago

n a call center the number of received calls in a day can be modeledby a Poisson random variable. We know that, on average, 0.5%

guapka [62]

Answer:Pois(ln(200))

Step-byy-step explanation:

Let N be the number of received calls in a day

That is

N∼Pois(λ).

0.5% = 0.5/100 = 1/200 of no calls

P(N=0)=e^−λ=1/200

-λ=e^(1/200)

λ=in(200)

Our number of calls in a day is distributed Pois(ln(200)).

7 0

3 years ago

Read 2 more answers

Solve the initial-value problem<br><br> y' = x^4 - \frac{1}{x}y, y(1) = 1.

natta225 [31]

The ODE is linear:

$y'=x^4-\dfrac yx$

$y'+\dfrac yx=x^4$

Multiplying both sides by $x$ gives

$xy'+y=x^5$

Notice that the left side can be condensed as the derivative of a product:

$(xy)'=x^5$

Integrating both sides with respect to $x$ yields

$xy=\dfrac{x^6}6+C$

$\implies y(x)=\dfrac{x^5}6+\dfrac Cx$

Since $y(1)=1$ ,

$1=\dfrac16+C\implies C=\dfrac56$

so that

$\boxed{y(x)=\dfrac{x^5}6+\dfrac5{6x}}$

4 0

3 years ago

A new neighborhood activity complex is being built in erie. The perimeter of the rectangular playing field is 256 yards. The len

maw [93]

Answer:

The length of the rectangular playing field is 84 yards and the width is 44 yards

Step-by-step explanation:

Let

x ------> the length of the rectangular playing field

y -----> the width of the rectangular playing field

we know that

The perimeter of the rectangular playing field is equal to

$P=2(x+y)$

$P=256\ yd$

$256=2(x+y)$ ------> equation A

we have that

$x=2y-4$ -----> equation B

Solve the system by substitution

Substitute equation B in equation A and solve for y

$256=2(2y-4+y)$

$128=3y-4$

$3y=128+4$

$3y=132$

$y=44$

Find the value of x

$x=2(44)-4=84$

therefore

The length of the rectangular playing field is 84 yards and the width is 44 yards

4 0

3 years ago