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vazorg [7]
3 years ago
13

A store carries four brands of DVD​ players, J,​ G, P and S. From past​ records, the manager found that the relative frequency o

f brand choice among customers varied. Using the given probability values for each of the four​ brands, find the probability that a random customer will choose brand J or brand P.
Mathematics
1 answer:
mr_godi [17]3 years ago
4 0

Answer:

The answer is below

Step-by-step explanation:

A store carries four brands of DVD​ players, J,​ G, P and S. From past​ records, the manager found that the relative frequency of brand choice among customers varied. Using the given probability values for each of the four​ brands, find the probability that a random customer will choose brand J or brand P.

P(J)=0.22​, ​P(G)=0.18​, ​P(P)=0.35​, ​P(S)=0.25

Answer: Probability is the ration of possible outcomes to the total number of possible outcomes. The probability of mutually exclusive events i.e. events that cannot occur at the same time is the sum of their individual probabilities. If two events A and B are mutually exclusive events, then:

P(A or B) = P(A) + P(B)

Given that P(J)=0.22​, ​P(G)=0.18​, ​P(P)=0.35​, ​P(S)=0.25, the probability that a random customer will choose brand J or brand P is given by:

P(J or P) = P(J) + P(P) = 0.22 + 0.35 = 0.57

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x = 1

Step-by-step explanation:

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That is equation of symmetry is x = 1

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Samuel, a county social worker, helps county residents who are struggling with different issues. He uses “c” to indicate client
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Which equation represents this sentence?
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Identify the three similar right triangles in the given diagram.
irga5000 [103]

Answer:

B. ΔABD, ΔADC, ΔDBC

Step-by-step explanation

Step -1 In ΔABD and ΔADC (from figure).

∠DAB=∠CAD (common in both triangles) ,

∠DBA=∠CDA =90 degree, and

∠BDA=∠DCA (rest angle of the two triangles).

therefore ΔABD similar to ΔADC (by AAA similarity theorem).

Step -2 In ΔDBC and ΔADC (from figure).

∠DCB=∠ACD (common in both triangles) ,

∠DBC=∠ADC =90 degree, and

∠CDB=∠CAD (rest angle of the two triangles).

therefore ΔDBC similar to ΔADC (by AAA similarity theorem).

Step -3 In ΔABD and ΔDBC (from figure).

∠BDA=∠BCD (because , ∠ACD=ADB from stap-1 and ∠ACD=∠BCD from figure) ,

∠DBA=∠CBD =90 degree, and

∠BAC=∠BDC (rest angle of the two triangles).

therefore ΔABD similar to ΔDBC (by AAA similarity theorem).

In the above step- ΔABD similar to ΔADC, ΔDBC similar to ΔADC and ΔABD similar to ΔDBC.

Hence ΔABD, ΔADC, ΔDBC similar to each other in the given figure.

4 0
2 years ago
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