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Stella [2.4K]
4 years ago
14

The probabilities of landing on 1, on 2, on 3, on 4, on 5 or on 6 when rolling a biased dice are shown in the table below.

Mathematics
1 answer:
ira [324]4 years ago
8 0

Answer:

99

Step-by-step explanation:

P(3) = 1-0.88 = 0.12

P(3 or 4) = 0.12 + 0.21 = 0.33

Expected no. = no

= 300(0.33) = 99

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PLEASE HELP ASAP!!! I NEED CORRECT ANSWERS ONLY PLEASE!!! I NEED TO FINISH THESE QUESTIONS BEFORE MIDNIGHT TONIGHT.
Natali [406]

Yo sup??

To solve this question we have to apply trigonometric ratios

sin31=UT/TV

TV=UT/sin31

=7.76

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Hope this helps.

6 0
3 years ago
Use the equation s=5q+8 to find the value of s when q=1.
Norma-Jean [14]

Answer:

s=13

Step-by-step explanation:

s=5q+8

plug in numbers

s=5(1)+8

solve 5*1

s=5+8

solve 5+8

s=13

7 0
3 years ago
in a pottery class, a teacher has 2/3 pound of clay for 6 students. If she gives each student a equal amount of clay, how much w
Jlenok [28]
She will give 1/9 pound of clay to each student
8 0
3 years ago
Read 2 more answers
Which property will Sarah use to solve 6x = 42?
Katarina [22]

Answer:

7

Step-by-step explanation:

4 0
4 years ago
(5) Find the Laplace transform of the following time functions: (a) f(t) = 20.5 + 10t + t 2 + δ(t), where δ(t) is the unit impul
Aloiza [94]

Answer

(a) F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}

(b) F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}

Step-by-step explanation:

(a) f(t) = 20.5 + 10t + t^2 + δ(t)

where δ(t) = unit impulse function

The Laplace transform of function f(t) is given as:

F(s) = \int\limits^a_0 f(s)e^{-st} \, dt

where a = ∞

=>  F(s) = \int\limits^a_0 {(20.5 + 10t + t^2 + d(t))e^{-st} \, dt

where d(t) = δ(t)

=> F(s) = \int\limits^a_0 {(20.5e^{-st} + 10te^{-st} + t^2e^{-st} + d(t)e^{-st}) \, dt

Integrating, we have:

=> F(s) = (20.5\frac{e^{-st}}{s} - 10\frac{(t + 1)e^{-st}}{s^2} - \frac{(st(st + 2) + 2)e^{-st}}{s^3}  )\left \{ {{a} \atop {0}} \right.

Inputting the boundary conditions t = a = ∞, t = 0:

F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}

(b) f(t) = e^{-t} + 4e^{-4t} + te^{-3t}

The Laplace transform of function f(t) is given as:

F(s) = \int\limits^a_0 (e^{-t} + 4e^{-4t} + te^{-3t} )e^{-st} \, dt

F(s) = \int\limits^a_0 (e^{-t}e^{-st} + 4e^{-4t}e^{-st} + te^{-3t}e^{-st} ) \, dt

F(s) = \int\limits^a_0 (e^{-t(1 + s)} + 4e^{-t(4 + s)} + te^{-t(3 + s)} ) \, dt

Integrating, we have:

F(s) = [\frac{-e^{-(s + 1)t}} {s + 1} - \frac{4e^{-(s + 4)}}{s + 4} - \frac{(3(s + 1)t + 1)e^{-3(s + 1)t})}{9(s + 1)^2}] \left \{ {{a} \atop {0}} \right.

Inputting the boundary condition, t = a = ∞, t = 0:

F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}

3 0
3 years ago
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