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elena-14-01-66 [18.8K]
3 years ago
7

Bryce randomly chooses a card from a standard deck of 52 containing 13 of each suit. He records the suit of each card in the tal

ly chart at the right.What are the experimental probabilities that Bryce chooses each suit? How do these compare to the theoretical probabilities?
clubs 6
diamonds 8
hearts 8
spades 10
Mathematics
1 answer:
Nikolay [14]3 years ago
3 0

Answer:

52 cards  / 4 suits  = 13 cards of each suit.

Theoretically picking a heart would be 13/52 = 1/4 probability.

Experimentally she picked 15 hearts out of 80 total tries. for a 15/80 = 3/16 probability, which is less than the theoretical probability.

1/4 - 3/16 = 1/16

The answer is A.

Step-by-step explanation:

The right answer is A - The theoretical probability of choosing a heart is StartFraction 1 over 16 EndFraction greater than the experimental probability of choosing a heart

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That would mean that each of the other 2 angles must add to 90 which makes each of them 45.

But the question doesn't allow that. B has to be greater than 90 which means that the other two angles must be less that 45 each.

the only answer that does that is A

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Find the measures of the angles of an isosceles obtuse triangle with one angle that is 32 degrees.
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32 degrees, 32 degrees and 116 degrees

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Help me please I’m stuck on this
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work for a publishing company. The company wants to send two employees to a statistics conference. To be​ fair, the company deci
Yuki888 [10]

Answer:

(a) S = {MR, MJ, MD, MC, RJ, RD, RC, JD, JC, DC}

(b) The probability that Roberto and John attend the​ conference is 0.10.

(c) The probability that Clarice attends the​ conference is 0.40.

(d) The probability that John stays​ home is 0.60.

Step-by-step explanation:

It is provided that :

Marco (<em>M</em>), Roberto (<em>R</em>), John (<em>J</em>), Dominique (<em>D</em>) and Clarice (<em>C</em>) works for the company.

The company selects two employees randomly to attend a statistics conference.

(a)

There are 5 employees from which the company has to select two employees to send to the conference.

So the total number of ways to select two employees is:

{5\choose 2}=\frac{5!}{2!(5-2)!}=\frac{5\times 4\times 3!}{2\times 3!}=10

The 10 possible samples are:

MR, MJ, MD, MC, RJ, RD, RC, JD, JC, DC

(b)

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

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

Here,

n (E) = favorable outcomes

N = Total number of outcomes.

The variable representing the selection of  Roberto and John is, <em>RJ</em>.

The favorable number of outcomes to select Roberto and John is, 1.

The total number of outcomes to select 2 employees is 10.

Compute the probability that Roberto and John attend the​ conference as follows:

P(RJ)=\frac{n(RJ)}{N}=\frac{1}{10}=0.10

Thus, the probability that Roberto and John attend the​ conference is 0.10.

(c)

The favorable outcomes of the event where Clarice attends the conference are:

n (C) = {MC, RC, JC and DC} = 4

Compute the probability that Clarice attends the​ conference as follows:

P(C)=\frac{n(C)}{N}=\frac{4}{10}=0.40

Thus, the probability that Clarice attends the​ conference is 0.40.

(d)

The favorable outcomes of the event where John does not attends the conference are:

n (J') = MR, MD, MC, RD, RC, DC

Compute the probability that John stays​ home as follows:

P(J')=\frac{n(J')}{N}=\frac{6}{10}=0.60

Thus, the probability that John stays​ home is 0.60.

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3 years ago
Add write your answer in simplest form 5 2/3 + 3 3/4
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Nine wholes with five twelves

9 5/12
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