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3 years ago

12

It took Jason 3 minutes and 48 seconds to complete a practice page in class. How many total seconds did Jason take to complete t

he practice page?

2 answers:

mr_godi [17]3 years ago

8 0

For this case we have that by definition, one minute equals 60 seconds. Then, we make the conversion:

$\frac {60 \ s} {1 \ m} * 3 \ m = 180 \ s$

Thus, 3 minutes equals 180 seconds.

Now:

$180 + 48 = 228$

So, it took Jason 228 seconds to complete a practice page in class.

Answer:

228 seconds

Nesterboy [21]3 years ago

5 0

Answer: 228

Step-by-step explanation:

3 X 60 = 180

180 + 48 = 228

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Consider randomly selecting a student at a certain university, and let A denote the event that the selected individual

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Answer:

(a) P( A ∩ B )=0.5 is not possible.

(b) 0.7

(c) 0.3

(d) 0.3

(e) 0.4

Step-by-step explanation:

Given information: The alphabet A and B represents the following events

A : Individual has a Visa credit card.

B: Individual has a MasterCard.

P(A)= 0.6 and P(B)=0.4.

(a)

We need to check whether P( A ∩ B ) can be 0.5 or not.

$A\cap B\subset A$ and $A\cap B\subset B$

$P(A\cap B)\leq P(A)$ and $P(A\cap B)\leq P(B)$

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From these two inequalities we conclude that

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Therefore, P( A ∩ B )=0.5 is not possible.

(b)

Let $P(A\cap B)=0.3$

We need to find the probability that student has one of these two types of cards.

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

Substitute the given values.

$P(A\cup B)=0.6+0.4-0.3=0.7$

Therefore the probability that student has one of these two types of cards is 0.7.

(c)

We need to find the probability that the selected student has neither type of card.

$P(A'\cup B')=1-P(A\cup B)$

$P(A'\cup B')=1-0.7=0.3$

Therefore the probability that the selected student has neither type of card is 0.3.

(d)

The event that the select student has a visa card, but not a mastercard is defined as

$A-B$

It can also written as

$A\cap B'$

The probability of this event is

$P(A\cap B')=P(A)-P(A\cap B)$

$P(A\cap B')=0.6-0.3=0.3$

Therefore the probability that the select student has a visa card, but not a mastercard is 0.3.

(e)

We need to find the probability that the selected student has exactly one of the two types of cards.

$P(A\cap B')+P(A\cap B')=P(A\cup B)-P(A\cap B)$

$P(A\cap B')+P(A\cap B')=0.7-0.3$

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Therefore the probability that the selected student has exactly one of the two types of cards is 0.4.

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Answer

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3 years ago