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Genrish500 [490]
3 years ago
14

Blocks numbered 0 through 9 are placed in a box, and a block is randomly picked.

Mathematics
2 answers:
aliya0001 [1]3 years ago
3 0
<span>Blocks numbered 0 through 9 are placed in a box, and a block is randomly picked.=3/10

</span><span>The probability of picking an odd prime number is . The probability of picking a number greater than 0 that is also a perfect square is=3/10</span>
faust18 [17]3 years ago
3 0

Answer:  2/5 and 3/10

Step-by-step explanation: Sample space S={0,1,2,3,4,5,6,7,8,9}

A:event of picking a odd prime number={1,3,5,7}

P(A)=4/10

      =2/5

B: event of picking a number greater 0 that is also a perfect square={1,4,9}

P(B)=3/10

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g Explain the difference between explanatory and response variables. Is it important to distinguish between the two in linear re
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2 years ago
An analysis of sales records for the last 120 weeks gives the following results. Assuming that these past data are a reliable gu
Mila [183]

Answer:

(a) 0.5333

(b) 0.6583

(c) 0.5583

(d) 0.7083

(e) 0.6167

Step-by-step explanation:

Denote the events as follows:

<em>A</em> = a competitor will advertise

<em>NA </em>= a competitor will not advertise

<em>L </em>= Low sales will be achieved

<em>M </em>= Medium sales will be achieved

<em>H </em>= High sales will be achieved

The data provided is of the form:

          Low (L)    Medium (M)    High (H)    Total    

A            32                   14                 18               64

NA          21                   12                 23              56

Total      53                  26                 41              120

The probability of an event <em>E</em> is:

P(E)=\frac{n(E)}{N}

n (E) = favorable outcomes of event <em>E</em>

N = Total number of outcomes

(a)

Compute the probability that next week the competitor will advertise as follows:

P(A)=\frac{n(A)}{N}=\frac{64}{120}=0.5333

Thus, the probability that next week the competitor will advertise is 0.5333.

(b)

Compute the probability that next week sales will not be high as follows:

P(H^{c})=1-P(H)=1-\frac{n(H)}{N}=1-\frac{41}{120}=\frac{120-41}{120}=0.6583

Thus, the probability that next week sales will not be high is 0.6583.

(c)

The events of achieving a medium or high sales are mutually exclusive.

Since the sales achieved will either be medium or high. They cannot be both.

So, P (M ∩ H) = 0.

Compute the probability that next week there will be medium or high sales will be achieved as follows:

P(M\cup H)=P(M)+P(H)=\frac{26}{120}+\frac{41}{120}=\frac{26+41}{120}=0.5583

Thus, the probability that next week there will be medium or high sales will be achieved is 0.5583.

(d)

Compute the probability that next week either the competitor will advertise, or only low sales will be achieved as follows:

P(A\cup L)=P(A)+P(L)-P(A\cap L)=\frac{64}{120}+\frac{53}{120}-\frac{32}{120}=\frac{85}{120}=0.7083

Thus, the  the probability that next week either the competitor will advertise, or only low sales will be achieved is 0.7083.

(e)

Compute the probability that next week either the competitor will not advertise, or high sales will be achieved as follows:

P(NA\cup H)=P(NA)+P(H)-P(NA\cap H)=\frac{56}{120}+\frac{41}{120}-\frac{23}{120}=\frac{74}{120}=0.6167

Thus, the  the probability that next week either the competitor will not advertise, or high sales will be achieved is 0.6167.

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