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vichka [17]
3 years ago
8

What is the correct listed sample space of total outcomes for rolling a die and flipping a coin?

Mathematics
1 answer:
RUDIKE [14]3 years ago
5 0
The answer would be C, because their are 12 possible outcomes. Each number on a dice can have 2 outcomes, and there are 6 numbers on a dice. Therefore, the answer is C.
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F(x) = 741
Marysya12 [62]

I think its option D

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3 years ago
Is this function linear or nonlinear? *
Bogdan [553]

Answer:

It's linear

Step-by-step explanation:

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3 years ago
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From the diagram below, if the sides AD = 3 and DC = 27, and BD = X + 3, find x.
Colt1911 [192]

Given:

• AD = 3

,

• DC = 27

,

• BD = x + 3

Let's solve for x.

To solve for x, apply the altitude formula:

\frac{AD}{BD}=\frac{BD}{DC}

Where BD is the altitude.

Cross multiply:

BD^2=AD*DC

Plug in the values and solve for x:

\begin{gathered} (x+3)^2=3*27 \\  \\ (x+3)^2=81 \end{gathered}

Take the square root of both sides:

\begin{gathered} \sqrt{(x+3)^2}=\sqrt{81} \\  \\ x+3=9 \\  \\ \text{ Subtract 3 from both sides:} \\ x+3-3=9-3 \\  \\ x=6 \end{gathered}

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ANSWER:

d. 6

4 0
1 year ago
Which of the following graphs is the graph of a linear function?
ziro4ka [17]

Answer:

C

Step-by-step explanation:

A and B are not straight lines, so they are not linear functions.

C and D are straight lines, so they are linear. However, D fails the vertical line test, so it is not a function.

3 0
3 years ago
A local grocery store chain has a sushi stand. That sushi stand makes samples for store customers to entice customers to purchas
Liula [17]

Answer: 0.99822

Step-by-step explanation:

Our inter-arrival time follows Poisson distribution with parameter 14 which is in hour so first we have to calculate this in minutes as we have to calculate probability in minutes.

So converting 14 into minutes will give  \frac{14}{60}=0.233

Let X=inter-arrival time between two customers

and here \lambda = 0.233

Probability(X is less than or equal to 2 minutes) = P(X=0) + P(X=1) + P(X=2)

Now the Poisson distribution has PDF = \frac{\exp ^{-\lambda }\times \lambda ^{X}}{X!}

So P(X = 0) = \frac{\exp ^{-0.233}\times 0.233^{0}}{0!} = 0.79215

P(X= 1) = \frac{\exp ^{-0.233}\times 0.233^{1}}{1!} = 0.18457

P(X=2) = \frac{\exp ^{-0.233}\times 0.233^{2}}{2!} = 0.02150

Now adding all three probability gives =  0.79215 +0.18457 + 0.02150=0.99822

3 0
3 years ago
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