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olya-2409 [2.1K]
3 years ago
15

ayuda !! Es la representación gráfica de una variable cuantitativa continua que agrupa intervalos de clases y es representada en

forma de barras a) Histograma b) Ojiva c) Polígono de frecuencia
Mathematics
1 answer:
nekit [7.7K]3 years ago
6 0

Answer:

<h2>a) Histograma (histogram).</h2>

Step-by-step explanation:

First of all, a continuous variable only admits whole numbers, not decimals.

Now, in statitics, there are several way to reprepsent some data collected from a survey. The most popular tool are bar-graphs or histograms, they are used to show frequency distributions through graphs, where we use interval classes.

Therefore, the right answer is a.

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Answer:

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Step-by-step explanation:

6 0
3 years ago
A new security system needs to be evaluated in the airport. The probability of a person being a security hazard is 4%. At the ch
ivolga24 [154]

Answer:

(a) 0.9412

(b) 0.9996 ≈ 1

Step-by-step explanation:

Denote the events a follows:

P = a person passes the security system

H = a person is a security hazard

Given:

P (H) = 0.04,\ P(P^{c}|H^{c})=0.02\ and\ P(P|H)=0.01

Then,

P(H^{c})=1-P(H)=1-0.04=0.96\\P(P|H^{c})=1-P(P|H)=1-0.02=0.98\\

(a)

Compute the probability that a person passes the security system using the total probability rule as follows:

The total probability rule states that: P(A)=P(A|B)P(B)+P(A|B^{c})P(B^{c})

The value of P (P) is:

P(P)=P(P|H)P(H)+P(P|H^{c})P(H^{c})\\=(0.01\times0.04)+(0.98\times0.96)\\=0.9412

Thus, the probability that a person passes the security system is 0.9412.

(b)

Compute the probability that a person who passes through the system is without any security problems as follows:

P(H^{c}|P)=\frac{P(P|H^{c})P(H^{c})}{P(P)} \\=\frac{0.98\times0.96}{0.9412} \\=0.9996\\\approx1

Thus, the probability that a person who passes through the system is without any security problems is approximately 1.

7 0
3 years ago
O DATA ANALYSIS AND STATISTICS Outcomes and event probability A number cube is rolled three times. An outcome is represented by
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ANSWER

EXPLANATION

For each of the three events, there are 8 possible outcomes, so the probability of each event is,

P(event)=\frac{favorable\text{ }outcomes}{possible\text{ }outcomes}=\frac{favorable\text{ }outcomes}{8}

• Event A:, there must be only odd numbers in the last two rolls, which are either ,EOO or OOO,. This event has 2 favorable outcomes, ,so its probability is 1/4,.

,

• Event B,: this is the complementary event to event A. Here the last two rolls must be even numbers - both or only one of the last two: ,OOE, EEO, EOE, OEE, EEE, OEO,. This event has 6 favorable outcomes, so ,its probability is 3/4,.

,

• Event C,: this event is when the first and last rolls are even numbers in the same event. This is either ,EOE or EEE,. Since this event has 2 favorable outcomes, ,its probability is 1/4,.

5 0
1 year ago
Write 5/6 as a decimal
Degger [83]

Answer:

0.83333333

Step-by-step explanation:

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3 0
3 years ago
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Explain why a set of points that defines a plane cannot be collinear
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Answer:

Three non-collinear points determine a plane.

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3 years ago
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