Please help me with this math question

Mathematics

1 answer:

Neko [114]3 years ago

3 0

Answer:

C .56

Step-by-step explanation:

well thier are 5 days that are less then 3 days and theirs 9 days in total so you would do 5/9 to get the chance of it being less then 3 days.

5/6 = .555556

You might be interested in

The perpendicular line of x-6y=2, (2, 4) in slope intercept form

pentagon [3]

The perpendicular line to x-6y=2, and passing through (2, 4) is y=-6x+16

3 0

2 years ago

Select a counter-example that makes the conclusion false.

Softa [21]

A counter-example could be

There are more than 3 marbles in the bag, and the other ones are red.

Hope this helps!

4 0

3 years ago

The sum of a number and 6 ad a mathematical expression

Gennadij [26K]

X+6=y thats the answer i got

7 0

2 years ago

Consider the following initial value problem, in which an input of large amplitude and short duration has been idealized as a de

Ganezh [65]

Answer:

a. $\mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

b. $\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}$

Step-by-step explanation:

The initial value problem is given as:

$y' +y = 7+\delta (t-3) \\ \\ y(0)=0$

Applying laplace transformation on the expression $y' +y = 7+\delta (t-3)$

to get $L[{y+y'} ]= L[{7 + \delta (t-3)}]$

$l\{y' \} + L \{y\} = L \{7\} + L \{ \delta (t-3\} \\ \\ sY(s) -y(0) +Y(s) = \dfrac{7}{s}+ e ^{-3s} \\ \\ (s+1) Y(s) -0 = \dfrac{7}{s}+ e^{-3s} \\ \\ \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

Taking inverse of Laplace transformation

$y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]$