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zloy xaker
2 months ago
13

Rework problem 4 from section 3.2 of your text, involving sets E and F. Suppose for this problem that Pr[E]=1/12, Pr[F]=1/12, an

d Pr[E∩F′]=0/1
1) What is Pr[E|F]?
2) What is Pr[F|E]?

Mathematics
1 answer:
Inessa [12.5K]2 months ago
3 0
The corresponding Venn Diagram for this issue is depicted below

P(E|F) and P(F|E) denote the conditional probabilities.

P(E|F) is calculated as P(E∩F) ÷ P(F) = ¹/₂ ÷ ¹/₂ = 1
P(F|E) is determined by P(F∩E) ÷ P(E) = ¹/₂ ÷ ¹/₂ = 1

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1199.28856995 to the nearest ten
lawyer [12517]
Hey! Hi! Hola! Aloha!


When rounding 1199.28856995 to the nearest tenth, the result is 119.3.

I hope this is helpful!
4 0
3 months ago
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A restaurant purchased kitchen equipment on January​ 1, 2017. On January​ 1, 2019, the value of the equipment was ​$14 comma 550
babunello [11817]

Answer:

\frac{dV(t)}{dt} = - 1675.38

Step-by-step explanation:

In 2017, the kitchen equipment's value was recorded at $14,550.

V(0)=$14550

Its following value is represented by V(t)=14550e^{-0.158t.

We need to ascertain the rate of value change as of January 1, 2019.

V(t)=14550e^{-0.158t

\frac{dV(t)}{dt} =\frac{d}{dt}14550e^{-0.158t

\frac{dV(t)}{dt} =14550 \frac{d}{dt}e^{-0.158t

\\Let u= -0.158t,\frac{du}{dt}=-0.158

\frac{dV(t)}{dt} =14550 \frac{d}{du}e^u\frac{du}{dt}

\frac{dV(t)}{dt} =14550 X -0.158 e^{-0.158t}=-2298.9e^{-0.158t}

As of 2019, which is 2 years later, we set t=2.

The rate of value change is

\frac{dV(t)}{dt} =-2298.9e^{-0.158X2}

=\frac{dV(t)}{dt} =-2298.9e^{-0.316}= -1675.38

3 0
3 months ago
Caitlyn read a book. She started reading at 9:45 am and ended at 10:37 am How many minutes did Caitlyn read
Zina [12379]

Response: 47 minutes

Detailed explanation:

5 0
2 months ago
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Claire is a manager at a toy packaging company. The company packs 80 boxes of toys every hour for the first 3 hours of the day.
Svet_ta [12734]

Response:

Refer to the figure attached.

Step-by-step explanation:

Each portion is treated as a function.

Let x represent the hours worked in a day

The company packs 80 boxes of toys each hour during the first three hours. Thus, y₁ = 80x, for x ∈ [0,3]

After three hours, the total boxes packed = 80 * 3 = 240

They pause packaging for two hours for a training session. Therefore, y₂ = 240, for x ∈ [3,5]

Next, for the subsequent four hours, they pack 20 boxes of toys hourly

y₃ = 20x + c, for x ∈ [5,9]

To determine c, equate y₂ and y₃ at x = 5

∴ 240 = 20 * 5 + c

∴ c = 240 - 20 * 5 = 240 - 100 = 140

∴ y₃ = 20x + 140, for x ∈ [5,9]

Consequently,

y₁ = 80x, for x ∈ [0,3]

y₂ = 240, for x ∈ [3,5]

y₃ = 20x + 140, for x ∈ [5,9]

The graph plotting the piecewise function that depicts the described situation is shown in the attached illustration.

8 0
3 months ago
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