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kotykmax [81]
3 years ago
8

The spinner below is spun then the a month of the year is randomly selected. Find each probability

Mathematics
2 answers:
bulgar [2K]3 years ago
7 0

Answer:

Step-by-step explanation:

P (blue)  fraction = 2/8   decimal=.25     percent=25%

P(purple) fraction=0/8   decimal=0         percent =0%

P(not green)  fraction=5/8     decimal=.625     percent= 62.5%

P(not brown) fraction=8/8     decimal=1   percent =100%

2)Refer to the table on air travel selected airports. Suppose a flight that arrived at <u>El Centro</u> is selected at random. What is the probability that the flight <u>did not arrive on time?</u> write your answer as a fraction, decimal and percent.    

80% on time so 20% are not on time

20/100 ÷ 20 =1/5

20%, .2, 1/5

3)The spinner has 8 sectors labeled with letters. Each sector of the spinner below is the same size, as shown.

If the arrow is spun 400 times , how many times  would it be  expected to land on a sector labeled B?

\frac{B}{Total} \\\frac{3}{8} =\frac{x}{400\\}

A)50                 C)150

B)100                D)250

PLEASE MARK ME AS BRAINLIEST

Kryger [21]3 years ago
6 0

Answer:

I think the answer is 0.3

Step-by-step explanation:

Sorry if am wrong.

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Answer:

4

Step-by-step explanation:

The answer is four because 10% of the whole tray (40) is 4

Equation:  

.1 * 40 = 4

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3 years ago
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The random variable X is exponentially distributed, where X represents the waiting time to be seated at a restaurant during the
erastova [34]

Answer:

The probability that the wait time is greater than 14 minutes  is 0.4786.

Step-by-step explanation:

The random variable <em>X</em> is defined as the waiting time to be seated at a restaurant during the evening.

The average waiting time is, <em>β</em> = 19 minutes.

The random variable <em>X</em> follows an Exponential distribution with parameter \lambda=\frac{1}{\beta}=\frac{1}{19}.

The probability distribution function of <em>X</em> is:

f(x)=\lambda e^{-\lambda x};\ x=0,1,2,3...

Compute the value of the event (<em>X</em> > 14) as follows:

P(X>14)=\int\limits^{\infty}_{14} {\lambda e^{-\lambda x}} \, dx=\lambda \int\limits^{\infty}_{14} {e^{-\lambda x}} \, dx\\=\lambda |\frac{e^{-\lambda x}}{-\lambda}|^{\infty}_{14}=e^{-\frac{1}{19} \times14}-0\\=0.4786

Thus, the probability that the wait time is greater than 14 minutes  is 0.4786.

7 0
2 years ago
Use the expression to answer the following questions. 5m – 8n + 2p – 12
Kobotan [32]

Answer:

12 is the constant

8 is the coefficient of the 2nd term

There are 4 terms

The terms are: 5m, 8n, 2p, and 12



7 0
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2 years ago
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